MathTest.uz
Algebra

Irratsional tengsizliklar

murakkab 210 daqiqa irratsional tengsizlikradical inequalityaniqlanish sohasiprincipal rootsquaringendpointpoleradikal kasrradikal ko‘paytmasubstitutionboundsverification

Nima uchun muhim?

Irratsional tengsizliklarda bir xil algebraik amal tenglama holatiga qaraganda ehtiyotkorroq ishlatiladi: darajaga ko‘tarishdan oldin ikki tomonning ishorasi, domain va strict/non-strict chegaralar tekshiriladi. Bu mavzu funksiyalar sohasi, grafiklar, parametrli masalalar va yuqori darajadagi algebraik tengsizliklarning tayanchi.

O'quv maqsadlari

  • Irratsional tengsizlikni tanish va real-domainni aniqlash
  • Juft indeksli radikal uchun radikand nonnegative shartini yozish
  • Radikal maxrajda bo‘lsa strict positive domainni qo‘llash
  • Principal square rootning nonnegative ekanidan tezkor xulosalar chiqarish
  • sqrt(A)<0, <=0, >0, >=0 master holatlarini qo‘llash
  • sqrt(A) ni musbat/manfiy konstant bilan solishtirishda case split qilish
  • sqrt(A)<c va sqrt(A)<=c ni xavfsiz square qilish
  • sqrt(A)>c va sqrt(A)>=c ni c ishorasiga qarab yechish
  • sqrt(A) va sqrt(B) ni common domain ostida radikandlar orqali taqqoslash
  • Toq ildiz monotonligini tengsizliklarda ishlatish
  • Ratsional radikandli tengsizliklarda numerator/denominator sign chart tuzish
  • Radikal ko‘paytma ishorasini algebraik factor va radical-zero endpointlar bilan tahlil qilish
  • Radikal maxrajli tengsizliklarda denominatorning musbatligini ishlatish
  • sqrt(A) ? B(x) tipida B ishorasiga qarab piecewise yechim qurish
  • t=sqrt(x) substitutionini inequality domaini bilan ishlatish
  • u=x^(1/4) substitutionini mixed-root tengsizliklarda qo‘llash
  • Nested radicals uchun ketma-ket domain shartlarini yozish
  • Ikki radikalli tengsizliklarni strukturaviy square/bound bilan yechish
  • Conjugate va monotonicity orqali root differencesni taqqoslash
  • Lower/upper bound orqali murakkab tengsizlikni qisqa isbotlash
  • Strict va non-strict endpoint hamda polelarni yakunda tekshirish
  • Source-bankdagi marked-answer va extraction xatolarini independent algebra bilan QA qilish
Irratsional tengsizlik — noma’lum radikal ichida qatnashadigan tengsizlik. Bu mavzuda eng muhim qoida: darajaga ko‘tarishdan oldin ikki tomonning ishorasini bilish kerak. Square funksiyasi [0,∞) da o‘suvchi, lekin butun real chiziqda one-to-one emas. Shu sabab sqrt(A)<B tengsizlikda B<=0 bo‘lsa yechim yo‘q, B>0 bo‘lsa esa 0<=A<B² ga o‘tish mumkin. Principal square root har doim nonnegative. Shuning uchun sqrt(A)>0 aynan A>0, sqrt(A)<=0 esa aynan A=0. Ikki principal root common domain ichida bir xil tartibni saqlaydi. Radikal maxraj esa alohida strict domain beradi: sqrt(Q) maxrajda bo‘lsa Q>0. Canonical workflow: original domainni yozish → qarshi tomon ishorasini aniqlash → xavfsiz equivalence yoki case split tanlash → algebraik tengsizlikni yechish → original domain bilan kesishish → strict endpoint, denominator pole va source-answer QA. Marked variant mathematical proof emas.

Ta'riflar

Irratsional tengsizlik · Radical inequality

Noma’lum kamida bitta radikal radikandida qatnashgan tengsizlik.

x ildiz ostida va taqqoslash belgisi mavjud.

Misol: $\sqrt{x+1}>2$

Bu emas: $\sqrt5+x>2$ da noma’lum radikal ichida emas.

💡 Real-domain birinchi qadam.

Aniqlanish sohasi · Domain

Tengsizlikning barcha ifodalari real va aniqlangan bo‘ladigan x lar to‘plami.

Yechim faqat original domain ichida bo‘lishi mumkin.

Misol: $\sqrt{x-3}$ uchun $x\ge3$.

Bu emas: $1/\sqrt{x-3}$ uchun $x\ge3$ emas, balki $x>3$.

💡 Final intersection majburiy.

Principal kvadrat ildiz · Principal square root

$\sqrt A$ — $A\ge0$ ning nonnegative square rooti.

Square root belgisi manfiy qiymat bermaydi.

Misol: $\sqrt9=3$.

Bu emas: $\sqrt9=-3$ noto‘g‘ri.

💡 Sign shortcutlarning asosi.

Strict tengsizlik · Strict inequality

< yoki > belgili tengsizlik; equality endpoint solutionga kirmaydi.

Chegara nuqtasida equality bo‘lsa endpoint ochiq.

Misol: $\sqrt x>2$ → $x>4$.

Bu emas: $x=4$ ni qo‘shish noto‘g‘ri.

💡 Polelar ham alohida chiqariladi.

Nonstrict tengsizlik · Non-strict inequality

≤ yoki ≥ belgili tengsizlik; equality admissible bo‘lsa endpoint kirishi mumkin.

Chegarani domain va denominator bilan tekshirib yopamiz.

Misol: $\sqrt x\ge2$ → $x\ge4$.

Bu emas: Denominator zero endpointni ≥ bo‘lsa ham kiritish mumkin emas.

💡 Endpoint QA.

Radikal maxraj · Radical denominator

Maxrajda principal radikal qatnashgan ifoda.

Maxraj real va zero bo‘lmasligi uchun radikand strict positive.

Misol: $1/\sqrt{x-2}$ uchun $x>2$.

Bu emas: $x=2$ domain emas.

💡 Strict domain.

Pole · Taqiqlangan nuqta

Kasr denominatori 0 bo‘ladigan x qiymati.

Sign chartda interval chegarasi, ammo solution emas.

Misol: $1/(x-4)$ da x=4 pole.

Bu emas: $[4,\infty)$ ko‘rinishida 4 ni kiritish noto‘g‘ri.

💡 Radikalli rational expressionsda juda muhim.

Monoton o‘suvchi funksiya · Increasing function

u<v bo‘lsa f(u)<f(v) tartibni saqlaydigan funksiya.

Square root [0,∞) da strictly increasing.

Misol: $A,B\ge0$: $\sqrt A<\sqrt B\iff A<B$.

Bu emas: Domain tashqarida radikandlarni shunchaki solishtirish mumkin emas.

💡 Inequality direction saqlanadi.

Toq ildiz monotonligi · Odd-root monotonicity

Toq n uchun x↦sqrt[n](x) butun real chiziqda strictly increasing.

Odd root inequalities direct radikand inequalityga tushadi.

Misol: $\sqrt[3]{3x-6}\le0\iff x\le2$.

Bu emas: Even root uchun negative radikandga ruxsat yo‘q.

💡 Sign restriction yo‘q.

Case split · Holatlarga ajratish

Qarshi tomonning ishorasi yoki algebraik factor signiga qarab solutionni holatlarga bo‘lish.

Square qilishdan oldin RHS manfiy yoki nonnegative ekanini ajratamiz.

Misol: $\sqrt A>B$: B<0 da domain avtomatik.

Bu emas: B ishorasini tekshirmay square qilish.

💡 Master technique.

Sign chart · Ishora jadvali

Factorlar critical pointlari bo‘yicha rational/algebraic expression ishorasini intervalma-interval tahlil qilish.

Numerator zero va denominator polelarni bir chiziqqa qo‘yamiz.

Misol: $\frac{x-2}{x-1}\ge0$.

Bu emas: Pole nuqtani solutionga qo‘shish.

💡 Rational radicandlar uchun asosiy vosita.

Radikal ko‘paytma · Radical product

Algebraik factor bilan nonnegative radical producti.

Radical positive bo‘lsa product signini algebraik factor belgilaydi; radical zero bo‘lsa product zero.

Misol: $(x-2)\sqrt{x+4}\le0$.

Bu emas: Domain tashqaridagi algebraic factor signini hisoblash yetarli emas.

💡 Zero-radical endpoint alohida.

Radikal kasr · Radical quotient

Numerator yoki denominatorida radikal qatnashgan kasrli tengsizlik.

Denominator sign/pole va radical domain birga boshqaradi.

Misol: $\frac{x-6}{\sqrt{x^2-8x+7}}\ge0$.

Bu emas: Denominator zero endpointni yopish.

💡 Original denominator restriction saqlanadi.

Almashtirish · Substitution

$t=\sqrt{x}$ yoki $u=x^{1/4}$ kabi nonnegative yangi o‘zgaruvchi kiritish.

Mixed radicals polynomial/rational inequalityga tushadi.

Misol: $t=\sqrt x\ge0$.

Bu emas: t ning manfiy intervalini x ga qaytarish.

💡 Constraintni saqlang.

Ichma-ich radikal · Nested radical

Bir radikalning ichida yana radikal qatnashgan ifoda.

Ichki va tashqi radikal domain shartlari birga bajarilishi kerak.

Misol: $\sqrt{7-2\sqrt{x-1}}$.

Bu emas: Faqat x-1≥0 ni yozish yetarli emas.

💡 Outer radicand ham ≥0.

Chegaraviy nuqta · Boundary point

Inequality algebraik equalityga aylangan critical x qiymati.

Strict/non-strict belgiga qarab inclusion tekshiriladi.

Misol: $\sqrt x<2$ da 4 boundary, lekin kirmaydi.

Bu emas: Har boundaryni avtomatik yopish.

💡 Domain bilan birga tekshiriladi.

Ekstremal baho · Bound

Ifodaning eng kichik yoki eng katta mumkin bo‘lgan qiymati orqali tengsizlikni isbotlash.

Uzoq squaring o‘rniga lower/upper bound ishlatamiz.

Misol: $\sqrt{x^2+25}\ge5$.

Bu emas: Taxminiy decimalni exact bound deb qabul qilish.

💡 No-solution/all-domain proof.

Qo‘shma ifoda · Conjugate

$\sqrt a-\sqrt b$ ni rationalized ko‘rinishga o‘tkazishda $\sqrt a+\sqrt b$ bilan bog‘liq juftlik.

Root differences monotonicityni ko‘rish osonlashadi.

Misol: $\sqrt t-\sqrt{t-1}=1/(\sqrt t+\sqrt{t-1})$.

Bu emas: Root difference doim ortadi deb taxmin qilish.

💡 Difference comparison.

Umumiy domain kesishmasi · Common domain

Bir nechta radikal/rational ifodaning barcha domain shartlari kesishmasi.

Har tomon alohida real bo‘lishi kerak.

Misol: $\sqrt{3x-10}>\sqrt{6-x}$ uchun $10/3\le x\le6$.

Bu emas: Faqat chap radikal domainini olish.

💡 Comparisondan oldin.

Source integrity QA · Manba yaxlitligi

Prompt, variant va marked-answerni independent algebra bilan tekshirish jarayoni.

Variant proof emas; endpoint va polelar source xatolarini tez fosh qiladi.

Misol: 8090 da canonical $(4,\infty)$, source `[4,∞)`.

Bu emas: Marked A ni avtomatik qabul qilish.

💡 Canonical content source’dan mustaqil.

Fundamental tushunchalar

Nonnegativity gate

Principal square root mavjud bo‘lsa u har doim nonnegative. Bu negative constant bilan taqqoslashni darhol hal qiladi.

$\sqrt A\ge0$

A≥0

Zero-level cases

sqrt(A)<=0 faqat sqrt(A)=0 bo‘lganda; sqrt(A)>0 esa aynan A>0 bo‘lganda bajariladi.

$\sqrt A\le0\iff A=0$

A real expression

Negative RHS shortcut

sqrt(A)>c yoki >=c da c<0 bo‘lsa har bir domain point yechim; sqrt(A)<c yoki <=c da c<0 bo‘lsa yechim yo‘q.

$c<0$

Original domain

Positive constant safe square

c>0 bo‘lsa root-vs-constant inequalities radikand inequalityga ekvivalent.

$\sqrt A

c>0

Endpoint semantics

<,> equality boundaryni chiqaradi; ≤,≥ esa point original domain/pole shartlarini bajarsa kiritadi.

Original expression defined

Equal-index root comparison

Common domain ichida principal square root strictly increasing; inequality direction o‘zgarmaydi.

$\sqrt A<\sqrt B\iff A

A,B≥0

Odd-root comparison

Odd root real line’da strictly increasing, shuning uchun sign va inequality direction to‘liq saqlanadi.

$\sqrt[3]A\le\sqrt[3]B\iff A\le B$

A,B real

Rational radicand domain

sqrt(P/Q) uchun P/Q≥0 va Q≠0; critical points P=0 va Q=0.

$\frac{P}{Q}\ge0$

Q≠0

Radical denominator strictness

1/sqrt(Q) mavjud bo‘lishi uchun Q>0. Equality Q=0 hech qachon endpoint sifatida kirmaydi.

$Q>0$

Radical denominator

Product sign decomposition

P(x)sqrt(Q(x)) da Q>0 regionda radical positive va sign=P sign; Q=0 bo‘lsa product zero.

$P\sqrt Q$

Q≥0

Quotient sign decomposition

P/sqrt(Q) da denominator Q>0 regionda strictly positive; inequality signini P boshqaradi.

$\frac{P}{\sqrt Q}$

Q>0

Variable RHS split

sqrt(A)>B kabi holatda B<0 region avtomatik, B≥0 regionda square qilinadi.

$\sqrt A>B$

A≥0

Variable RHS upper bound

sqrt(A)<B uchun B>0 majburiy; keyin A<B². B≤0 regionda solution yo‘q.

$\sqrt A

B>0, A≥0

sqrt(x) substitution

x=t² va t≥0 orqali sqrt(x) qatnashgan rational inequality oddiy sign chartga aylanadi.

$t=\sqrt x\ge0$

x≥0

Fourth-root substitution

sqrt(x)=u² va fourthroot(x)=u, u≥0. Mixed roots quadratic u inequalityga tushadi.

$u=x^{1/4}\ge0$

x≥0

Nested domain chain

Ichki radikalning domaini outer radikand inequalityga qo‘shiladi; barcha shartlar bir vaqtning o‘zida bajariladi.

Every nested radical real

Conjugate difference monotonicity

sqrt(t)-sqrt(t-a)=a/(sqrt(t)+sqrt(t-a)); denominator ortgani uchun difference kamayadi.

$\sqrt t-\sqrt{t-a}=\frac{a}{\sqrt t+\sqrt{t-a}}$

a>0, t≥a

Root-sum structural lower bound

A,B≥0 bo‘lsa sqrt(A)+sqrt(B)≥sqrt(A+B), equality iff AB=0.

$\sqrt A+\sqrt B\ge\sqrt{A+B}$

A,B≥0

Pythagorean comparison

A+B=C strukturada radical-sum vs sqrt(C) cross product signi bilan hal bo‘ladi.

$(\sqrt A+\sqrt B)^2=C+2\sqrt{AB}$

A,B≥0

Extremal minimum at x=0

sqrt(x²+a²) kabi even expressions x=0 da minimumga ega; weighted sums uchun tez bound beradi.

$\sqrt{x^2+a^2}\ge |a|$

a real

Cancellation preserves exclusions

Common factor cancel qilinganda original denominator zero nuqtasi qayta kiritilmaydi.

Original denominator nonzero

Strict inequality QA

Algebraik equality roots strict inequalityda ochiq boundary bo‘ladi; denominator poles har qanday belgida chiqariladi.

< or >

Source answer is not proof

Marked option canonical solutionga qarshi chiqsa domain, monotonicity va exact algebra ustun turadi.

Independent verification

Formula kutubxonasi

Square-root domain

$$\sqrt{A(x)}\text{ real}\Longleftrightarrow A(x)\ge0$$
  • $A(x)$ — radikand

Principal square root faqat nonnegative radikandda real.

Shart: Real sonlar sohasi

Xususiy holatlar: Maxrajda bo‘lsa strict A>0.

Radical denominator domain

$$\frac{1}{\sqrt{A(x)}}\text{ defined}\Longleftrightarrow A(x)>0$$
  • $A(x)$ — denominator radikandi

Radikand 0 bo‘lsa root 0 va kasr aniqlanmaydi.

Shart: Real sonlar; denominator nonzero

Xususiy holatlar: A=0 pole.

Square root negative strict

$$\sqrt A<0\Longleftrightarrow\varnothing$$

Principal root manfiy bo‘la olmaydi.

Shart: A≥0 domain

Xususiy holatlar: Har qanday real-defined square root.

Square root nonpositive

$$\sqrt A\le0\Longleftrightarrow A=0$$

Nonnegative root faqat 0 bo‘lganda ≤0.

Shart: A≥0

Xususiy holatlar: Strict <0 esa yechimsiz.

Square root positive

$$\sqrt A>0\Longleftrightarrow A>0$$

Root positive iff radikand positive.

Shart: Real square root

Xususiy holatlar: A=0 da equality 0.

Square root nonnegative

$$\sqrt A\ge0\Longleftrightarrow A\ge0$$

Tengsizlikning o‘zi aynan domainni beradi.

Shart: Real sonlar

Xususiy holatlar: Domain endpointlar kiradi.

Upper strict constant — impossible branch

$$c\le0\Longrightarrow \sqrt A
  • $c$ — constant

Nonnegative root nonpositive sondan kichik bo‘la olmaydi.

Shart: A≥0

Xususiy holatlar: c=0 ham yechimsiz.

Upper strict constant — positive branch

$$c>0:\quad \sqrt A
  • $c$ — positive constant

[0,∞) da square tartibni saqlaydi.

Shart: c>0

Xususiy holatlar: Strict boundary A=c² kirmaydi.

Upper nonstrict constant — negative branch

$$c<0\Longrightarrow \sqrt A\le c\text{ has no real solution}$$
  • $c$ — constant

Nonnegative root negative sondan ≤ bo‘la olmaydi.

Shart: A≥0

Xususiy holatlar: c=0 alohida A=0.

Upper nonstrict constant — nonnegative branch

$$c\ge0:\quad \sqrt A\le c\Longleftrightarrow 0\le A\le c^2$$
  • $c$ — nonnegative constant

Square qilish inequality directionni saqlaydi.

Shart: c≥0

Xususiy holatlar: c=0 → A=0.

Lower strict constant — negative branch

$$c<0:\quad \sqrt A>c\Longleftrightarrow A\ge0$$
  • $c$ — negative constant

Har real-defined principal root negative constantdan katta.

Shart: c<0

Xususiy holatlar: Original domainning o‘zi solution.

Lower strict constant — nonnegative branch

$$c\ge0:\quad \sqrt A>c\Longleftrightarrow A>c^2$$
  • $c$ — nonnegative constant

Strict order square orqali saqlanadi.

Shart: c≥0

Xususiy holatlar: c=0 → A>0.

Lower nonstrict constant — nonpositive branch

$$c\le0:\quad \sqrt A\ge c\Longleftrightarrow A\ge0$$
  • $c$ — nonpositive constant

Har real-defined root c dan kam emas.

Shart: c≤0

Xususiy holatlar: Domainning o‘zi solution.

Lower nonstrict constant — positive branch

$$c>0:\quad \sqrt A\ge c\Longleftrightarrow A\ge c^2$$
  • $c$ — positive constant

Nonstrict order square orqali saqlanadi.

Shart: c>0

Xususiy holatlar: Boundary A=c² kiradi.

Equal-index root comparison

$$A,B\ge0:\quad \sqrt A\;\square\;\sqrt B\Longleftrightarrow A\;\square\;B$$
  • $\square$ — <, ≤, > yoki ≥

Principal sqrt strictly increasing.

Shart: A,B≥0

Xususiy holatlar: Common domain majburiy.

Odd-root comparison

$$\sqrt[n]{A}\;\square\;\sqrt[n]{B}\Longleftrightarrow A\;\square\;B$$
  • $n$ — toq musbat integer

Odd-root butun real chiziqda strictly increasing.

Shart: n toq; A,B real

Xususiy holatlar: Negative radikandlar mumkin.

Variable RHS upper strict

$$\sqrt A0,\quad 0\le A
  • $B$ — variable RHS

RHS positive bo‘lishi shart; keyin safe square.

Shart: Real expressions

Xususiy holatlar: Strict B>0.

Variable RHS upper nonstrict

$$\sqrt A\le B\Longleftrightarrow B\ge0,\quad 0\le A\le B^2$$
  • $B$ — variable RHS

RHS nonnegative va squared inequality.

Shart: Real expressions

Xususiy holatlar: B=0 faqat A=0.

Variable RHS lower strict

$$\sqrt A>B\Longleftrightarrow (A\ge0,\ B<0)\;\text{or}\;(B\ge0,\ A>B^2)$$
  • $B$ — variable RHS

Negative RHS region automatic; nonnegative RHS region safe-square.

Shart: Real expressions

Xususiy holatlar: B=0 ikkinchi branchda A>0.

Variable RHS lower nonstrict

$$\sqrt A\ge B\Longleftrightarrow (A\ge0,\ B\le0)\;\text{or}\;(B>0,\ A\ge B^2)$$
  • $B$ — variable RHS

Nonpositive RHS region domainning o‘zi; positive RHS square qilinadi.

Shart: Real expressions

Xususiy holatlar: B=0 first branchga kiradi.

Rational radicand domain

$$\sqrt{\frac{P}{Q}}\text{ real}\Longleftrightarrow \frac{P}{Q}\ge0,\quad Q\ne0$$
  • $P,Q$ — real expressions

Rational expression nonnegative bo‘lishi kerak.

Shart: Q≠0

Xususiy holatlar: P=0 allowed; Q=0 never.

Radical quotient sign

$$Q>0:\quad \frac{P}{\sqrt Q}\;\square\;0\Longleftrightarrow P\;\square\;0$$
  • $\square$ — <,≤,>,≥

Denominator strictly positive, shuning uchun sign numeratorga teng.

Shart: Q>0

Xususiy holatlar: Q=0 excluded.

sqrt substitution

$$t=\sqrt x\ge0,\qquad x=t^2$$
  • $t$ — nonnegative substitute

Mixed x va sqrt(x) expressionlarni t ga o‘tkazadi.

Shart: x≥0

Xususiy holatlar: Back-substitute x=t².

Fourth-root substitution

$$u=x^{1/4}\ge0,\qquad \sqrt x=u^2,\quad x=u^4$$
  • $u$ — fourth-root substitute

sqrt(x) va fourthroot(x) aralashmasini quadratic u expressionga aylantiradi.

Shart: x≥0

Xususiy holatlar: u≥0 constraint.

Root-sum lower bound

$$\sqrt A+\sqrt B\ge\sqrt{A+B}$$

Cross term 2sqrt(AB) nonnegative.

Shart: A,B≥0

Xususiy holatlar: Equality iff AB=0.

Conjugate root difference

$$\sqrt t-\sqrt{t-a}=\frac{a}{\sqrt t+\sqrt{t-a}}$$
  • $a$ — positive gap

Difference denominator ortishi bilan kamayadi.

Shart: a>0, t≥a

Xususiy holatlar: Useful for monotonicity without calculus.

Teoremalar va isbotlar

📐 Square-root va konstantani taqqoslash teoremasi

Principal square rootni konstant bilan taqqoslashda constant ishorasi solution strukturasini to‘liq belgilaydi.

Root nonnegative; faqat ikkala tomon nonnegative bo‘lganda square orderni xavfsiz saqlaydi.

Isbotni ko'rsatish

Berilgan: A≥0 regionda principal sqrt(A) va real c.

Isbotlash kerak: c ishorasiga qarab sqrt(A) ? c master holatlarini ko‘rsatish.

  1. sqrt(A)≥0 har bir domain pointda.
  2. Agar c<0 bo‘lsa sqrt(A)>c va sqrt(A)≥c avtomatik; sqrt(A)<c va sqrt(A)≤c imkonsiz.
  3. c=0 holatda sqrt(A)<0 imkonsiz, sqrt(A)≤0 iff A=0, sqrt(A)>0 iff A>0, sqrt(A)≥0 domainning o‘zi.
  4. c>0 bo‘lsa ikkala tomon nonnegative; square strictly increasing, shuning uchun inequality aynan A ? c² ga o‘tadi.

Constant-comparison master cases isbotlandi. ∎

📐 Variable RHS sign-split teoremasi

sqrt(A) ? B(x) tipidagi tengsizlik B ishorasi bo‘yicha bo‘linib, nonnegative branchda square orqali ekvivalent algebraik tengsizlikka aylanadi.

Square qilishdan oldin qarshi tomon qaysi tomonda ekanini bilish kerak.

Isbotni ko'rsatish

Berilgan: sqrt(A) ? B(x), original domain A≥0.

Isbotlash kerak: B sign split formulasini isbotlash.

  1. sqrt(A) har doim ≥0.
  2. B<0 regionda sqrt(A)>B va ≥B avtomatik; <B va ≤B imkonsiz.
  3. B=0 region zero-level qoidalarga tushadi.
  4. B>0 regionda ikkala tomon nonnegative, shuning uchun square tartibni saqlaydi: sqrt(A)?B iff A?B².
  5. Regionlar original domain bilan kesishib birlashtiriladi.

Variable RHS piecewise equivalence exact. ∎

📐 Principal root monotonligi teoremasi

A,B≥0 common domainda sqrt(A) va sqrt(B) orasidagi <,≤,>,≥ tartib A va B orasidagi ayni tartibga ekvivalent.

Square root [0,∞) da strictly increasing.

Isbotni ko'rsatish

Berilgan: A≥0 va B≥0.

Isbotlash kerak: sqrt(A) va sqrt(B) tartibi radikandlar tartibiga aynan tengligini ko‘rsatish.

  1. 0≤A<B bo‘lsa strictly increasing sqrt sabab sqrt(A)<sqrt(B).
  2. A=B bo‘lsa principal roots teng.
  3. A>B bo‘lsa xuddi shu sabab sqrt(A)>sqrt(B).
  4. Nonstrict belgilar equality holatini qo‘shish orqali keladi.

Barcha to‘rtta comparison belgisi uchun ekvivalentlik isbotlandi. ∎

📐 Radikal product/quotient sign teoremasi

Principal radical nonnegative; positive radicandda strictly positive. Shu sabab product va positive radical denominatorli quotient signi algebraik factor/numerator signi bilan boshqariladi, radical-zero boundary alohida olinadi.

Radical sign noma’lum emas: u 0 yoki positive.

Isbotni ko'rsatish

Berilgan: P(x),Q(x) real expressions.

Isbotlash kerak: P sqrt(Q) va P/sqrt(Q) sign qoidalarini ko‘rsatish.

  1. Q<0 da original radical real emas.
  2. Q=0 da sqrt(Q)=0; product P sqrt(Q)=0, lekin quotient aniqlanmagan.
  3. Q>0 da sqrt(Q)>0.
  4. Positive songa ko‘paytirish yoki bo‘lish inequality directionni o‘zgartirmaydi; shuning uchun sign P bilan bir xil.

Product/quotient sign analysis domain bilan exact. ∎

📐 Root-sum structural bound teoremasi

A,B≥0 bo‘lsa sqrt(A)+sqrt(B)≥sqrt(A+B), equality aynan AB=0 bo‘lganda.

Square qilinganda yagona qo‘shimcha had 2sqrt(AB)≥0.

Isbotni ko'rsatish

Berilgan: A≥0 va B≥0.

Isbotlash kerak: sqrt(A)+sqrt(B)≥sqrt(A+B), equality iff AB=0.

  1. Chap tomonning kvadrati A+B+2sqrt(AB).
  2. O‘ng tomonning kvadrati A+B.
  3. 2sqrt(AB)≥0, demak chap kvadrat kam emas.
  4. Nonnegative sonlarda square orderni saqlaydi, shuning uchun original inequality keladi.
  5. Equality iff sqrt(AB)=0 iff AB=0.

Root-sum bound va equality condition isbotlandi. ∎

Yechilgan misollar

oson $\sqrt{x}<-14$ tengsizlikni yeching.

💡 Maslahat: Principal square rootning ishorasini o‘ylang.

  1. Domain x≥0.
  2. Har domain pointda sqrt(x)≥0.
  3. 0<-14 noto‘g‘ri, demak hech qanday x yo‘q.

✅ Javob: $\varnothing$

Nega bu usul ishlaydi: Negative RHS shortcut square qilishsiz masalani yopadi.

⚠️ Source 8072 bilan mos.

oson $\sqrt[3]{3x-6}\le0$ tengsizlikni yeching.

💡 Maslahat: Cube root strictly increasing.

  1. Cube-root butun real chiziqda o‘suvchi.
  2. 3x-6≤0.
  3. x≤2.

✅ Javob: $(-\infty,2]$

Nega bu usul ishlaydi: Odd-root comparison signni va inequality directionni saqlaydi.

⚠️ Source 8079 faqat {2} ni belgilagan; canonical interval x≤2.

oson $\sqrt{x-2}\le0$ tengsizlikni yeching.

💡 Maslahat: Square root nonnegative.

  1. Domain x≥2.
  2. Nonnegative root ≤0 faqat root=0 bo‘lganda.
  3. x-2=0, demak x=2.

✅ Javob: $\{2\}$

Nega bu usul ishlaydi: Zero-level master case ishlaydi.

⚠️ Source 8078 bilan mos.

ortacha $\sqrt{\frac{2x-2}{x-4}}\ge-1$ tengsizlikni yeching.

💡 Maslahat: RHS negative; faqat domainni toping.

  1. Principal root har doim ≥0, shuning uchun -1 dan katta yoki teng avtomatik.
  2. Radikand domain: 2(x-1)/(x-4)≥0, x≠4.
  3. Sign chart critical points 1 va 4 beradi.
  4. Domain (-∞,1]∪(4,∞).

✅ Javob: $(-\infty,1]\cup(4,\infty)$

Nega bu usul ishlaydi: Negative lower boundda solution original domainning o‘zi.

Muqobil usul: Sign chartni numerator/denominator critical pointlari bilan chizing.

⚠️ Source 8082 x=4 ni noto‘g‘ri kiritgan.

oson $\sqrt{x^2-4}>0$ tengsizlikni yeching.

💡 Maslahat: sqrt(A)>0 iff A>0.

  1. x²-4>0.
  2. (x-2)(x+2)>0.
  3. x<-2 yoki x>2.

✅ Javob: $(-\infty,-2)\cup(2,\infty)$

Nega bu usul ishlaydi: Strict positivity radikandning strict positivitysiga aynan teng.

Muqobil usul: Absolute value orqali |x|>2.

⚠️ Source 8084 x=2 endpointini noto‘g‘ri kiritgan.

oson $\sqrt{3x+5}\le0$ tengsizlikni yeching.

💡 Maslahat: Nonnegative root ≤0 faqat 0.

  1. 3x+5=0.
  2. x=-5/3.
  3. Bu pointda radikand 0 va root real.

✅ Javob: $\{-\frac53\}$

Nega bu usul ishlaydi: Zero-level nonstrict case.

Muqobil usul: Domain x≥-5/3 ichida faqat boundary equality.

⚠️ Source 8089 butun chap intervalni belgilagan; canonical faqat x=-5/3.

oson $\sqrt{x}>2$ tengsizlikni yeching.

💡 Maslahat: RHS positive, safe square.

  1. Domain x≥0.
  2. Ikkala tomon nonnegative, square: x>4.
  3. x>4 domain ichida.

✅ Javob: $(4,\infty)$

Nega bu usul ishlaydi: Positive constant bilan strict comparison square orqali ekvivalent.

Muqobil usul: sqrt monotonligi.

⚠️ Source 8090 x=4 ni noto‘g‘ri yopgan.

oson $\sqrt{3-2x}<2$ tengsizlikni yeching.

💡 Maslahat: Domain va positive RHSni birga ishlating.

  1. Domain 3-2x≥0 → x≤3/2.
  2. 2>0, square: 3-2x<4 → x>-1/2.
  3. Kesishma (-1/2,3/2].

✅ Javob: $(-\frac12,\frac32]$

Nega bu usul ishlaydi: Safe square + original domain intersection.

Muqobil usul: Intervalni ikki shart kesishmasi sifatida yozing.

⚠️ Source 8092 bilan mos.

oson $\sqrt{3x-6}<3$ tengsizlikni yeching.

💡 Maslahat: Strict upper bound endpointini tekshiring.

  1. Domain x≥2.
  2. Square: 3x-6<9 → x<5.
  3. Kesishma [2,5).

✅ Javob: $[2,5)$

Nega bu usul ishlaydi: Strict inequality equality boundary 5 ni chiqaradi.

Muqobil usul: 0≤3x-6<9 ni birgalikda yeching.

⚠️ Source 8093 x=5 ni noto‘g‘ri kiritgan.

ortacha $\sqrt{-0.5x-9}\ge5$ tengsizlikni yeching.

💡 Maslahat: Positive lower boundga square qo‘llang.

  1. 5>0, shuning uchun -0.5x-9≥25.
  2. -0.5x≥34.
  3. Negative songa bo‘lganda inequality aylanadi: x≤-68.
  4. Bu shart radikandni avtomatik nonnegative qiladi.

✅ Javob: $(-\infty,-68]$

Nega bu usul ishlaydi: Positive threshold master rule va negative coefficient sign reversal birga ishlaydi.

⚠️ Source 8096 bilan mos.

oson $\sqrt{4x-16}\le32$ tengsizlikni yeching.

💡 Maslahat: 0≤4x-16≤32².

  1. Domain x≥4.
  2. Square upper bound: 4x-16≤1024.
  3. 4x≤1040 → x≤260.
  4. Kesishma [4,260].

✅ Javob: $[4,260]$

Nega bu usul ishlaydi: Nonstrict positive upper bound endpointsni saqlaydi.

⚠️ Source 8097 bilan mos.

murakkab $\frac{\sqrt{x+2}}{5-3x}>\frac12$ tengsizlikni yeching.

💡 Maslahat: RHS positive bo‘lgani uchun denominator ishorasini avval aniqlang.

  1. Domain x≥-2 va x≠5/3.
  2. Agar 5-3x<0 bo‘lsa LHS≤0, 1/2 dan katta bo‘la olmaydi; demak x<5/3.
  3. Positive denominator bilan 2sqrt(x+2)>5-3x; RHS ham positive.
  4. Square: 4(x+2)>(5-3x)².
  5. Bu 9x²-34x+17<0 ga teng.
  6. Quadratic roots (17±2sqrt34)/9; interval ular orasida.
  7. x<5/3 bilan kesishma: ((17-2sqrt34)/9,5/3).

✅ Javob: $(\frac{17-2\sqrt{34}}9,\frac53)$

Nega bu usul ishlaydi: Positive RHS denominator signini majbur qiladi; keyin safe squaring.

Muqobil usul: Hammasini bir tomonga olib sign chart qilish mumkin.

⚠️ Source 8098 lower endpoint -3/7 deb bergan; noto‘g‘ri.

ortacha $\sqrt{\frac{x-2}{x-1}}\ge1$ tengsizlikni yeching.

💡 Maslahat: Square faqat radikand domainida.

  1. RHS 1 positive, square: (x-2)/(x-1)≥1, original denominator x≠1.
  2. Subtract 1: -1/(x-1)≥0.
  3. Bu x<1 ni beradi.
  4. x<1 da original radikand ham positive.

✅ Javob: $(-\infty,1)$

Nega bu usul ishlaydi: Squared rational inequality original domainni saqlaydi.

Muqobil usul: Radikand domainni alohida sign chart bilan tekshirish mumkin.

⚠️ Source 8099 x=1 poleni noto‘g‘ri kiritgan.

murakkab $\sqrt{\frac{3x+1}{x+2}}<1$ tengsizlikni yeching.

💡 Maslahat: Radikand domain va ratio<1 ni kesishiring.

  1. Domain (3x+1)/(x+2)≥0, x≠-2: (-∞,-2)∪[-1/3,∞).
  2. RHS positive, square: (3x+1)/(x+2)<1.
  3. Equivalent (2x-1)/(x+2)<0 → (-2,1/2).
  4. Domain bilan kesishma [-1/3,1/2).

✅ Javob: $[-\frac13,\frac12)$

Nega bu usul ishlaydi: Rational radicand domain va squared comparison ikki alohida sign chart beradi.

⚠️ Source 8101 variantlarining hech biri endpointlarni to‘g‘ri bermaydi.

ortacha $\sqrt{\frac{x-7}{3x-21}}<6$ tengsizlikni yeching.

💡 Maslahat: Radicandni avval soddalashtiring, lekin original pole saqlansin.

  1. 3x-21=3(x-7).
  2. x≠7 da radikand 1/3 ga teng.
  3. sqrt(1/3)<6 rost.
  4. Original x=7 da 0/0 undefined, shuning uchun chiqariladi.

✅ Javob: $(-\infty,7)\cup(7,\infty)$

Nega bu usul ishlaydi: Cancellation original denominator restrictionni bekor qilmaydi.

Muqobil usul: Original expressionga x=7 ni qo‘yib ko‘ring.

⚠️ Source 8103 x=7 ni noto‘g‘ri kiritgan.

oson $\sqrt{2x-x^2}\ge1$ tengsizlikni yeching.

💡 Maslahat: Positive lower boundga square.

  1. 2x-x²≥1.
  2. -(x-1)²≥0.
  3. Faqat (x-1)²=0 mumkin.
  4. x=1.

✅ Javob: $\{1\}$

Nega bu usul ishlaydi: Radikand threshold domainni avtomatik ta’minlaydi.

⚠️ Source 8105 bilan mos.

ortacha $\sqrt{4x^2-12x+41}\le2$ tengsizlikni yeching.

💡 Maslahat: Radikandning minimumini toping.

  1. 4x²-12x+41=4(x-3/2)²+32.
  2. Radikand har doim ≥32.
  3. Shuning uchun LHS≥sqrt32>2.
  4. Yechim yo‘q.

✅ Javob: $\varnothing$

Nega bu usul ishlaydi: Extremal lower bound uzoq quadratic inequalityni darhol yopadi.

Muqobil usul: Square qilsangiz 4x²-12x+37≤0 chiqadi va diskriminant manfiy.

⚠️ Source 8107 [1,5] deb belgilagan; noto‘g‘ri.

murakkab $\sqrt{x^2+3x-2}<2$ tengsizlikni yeching.

💡 Maslahat: 0≤radikand<4 sistemani yeching.

  1. Domain: x²+3x-2≥0. Uning roots r1=(-3-sqrt17)/2, r2=(-3+sqrt17)/2; domain (-∞,r1]∪[r2,∞).
  2. Upper strict condition: x²+3x-2<4 → x²+3x-6<0.
  3. Bu roots s1=(-3-sqrt33)/2, s2=(-3+sqrt33)/2 orasida bajariladi.
  4. Kesishma (s1,r1]∪[r2,s2).

✅ Javob: $(\frac{-3-\sqrt{33}}2,\frac{-3-\sqrt{17}}2]\cup[\frac{-3+\sqrt{17}}2,\frac{-3+\sqrt{33}}2)$

Nega bu usul ishlaydi: Square-root upper bound ikki radikand inequality: domain nonnegative va thresholddan kichik.

Muqobil usul: 0≤A<4 ni bitta compound condition deb ko‘ring.

⚠️ Source 8108 `(-5,2)` deb bergan; noto‘g‘ri.

ortacha $\sqrt{-x^2+6x-4}>1$ tengsizlikni yeching.

💡 Maslahat: Positive thresholdga square.

  1. Square: -x²+6x-4>1.
  2. -x²+6x-5>0.
  3. Equivalent (x-1)(x-5)<0.
  4. Demak 1<x<5.

✅ Javob: $(1,5)$

Nega bu usul ishlaydi: Strict lower bound A>1 automatically A≥0 ni ham ta’minlaydi.

⚠️ Source 8109 endpointlarni noto‘g‘ri yopgan.

oson $\sqrt{x+2}<\sqrt{x}$ tengsizlikni yeching.

💡 Maslahat: Common domainda radikandlarni solishtiring.

  1. Common domain x≥0.
  2. Square-root strictly increasing, shuning uchun x+2<x talab qilinadi.
  3. 2<0 imkonsiz.

✅ Javob: $\varnothing$

Nega bu usul ishlaydi: Equal-index root comparison bir qadamda contradiction beradi.

⚠️ Source 8110 bilan mos.

oson $\sqrt{x+3}>\sqrt{x}$ tengsizlikni yeching.

💡 Maslahat: Common domain va radikand comparison.

  1. Common domain x≥0.
  2. x+3>x barcha x uchun.
  3. Shuning uchun butun common domain solution.

✅ Javob: $[0,\infty)$

Nega bu usul ishlaydi: sqrt strictly increasing va radikand difference 3 positive.

⚠️ Source 8111 bilan mos.

ortacha $\sqrt{3x-10}>\sqrt{6-x}$ tengsizlikni yeching.

💡 Maslahat: Avval ikkala radikalning common domaini.

  1. 3x-10≥0 → x≥10/3.
  2. 6-x≥0 → x≤6.
  3. Common domain [10/3,6].
  4. Radikand comparison: 3x-10>6-x →4x>16 →x>4.
  5. Kesishma (4,6].

✅ Javob: $(4,6]$

Nega bu usul ishlaydi: Common domain ostida root comparison aynan radikand comparison.

Muqobil usul: Ikkala tomonni square qiling.

⚠️ Source 8114 x=4 ni noto‘g‘ri kiritgan.

ortacha $\sqrt{3x-4}<\sqrt{4-x}$ tengsizlikni yeching.

💡 Maslahat: Common domainni toping.

  1. 3x-4≥0 → x≥4/3.
  2. 4-x≥0 → x≤4.
  3. Common domain [4/3,4].
  4. 3x-4<4-x →4x<8 →x<2.
  5. Natija [4/3,2).

✅ Javob: $[\frac43,2)$

Nega bu usul ishlaydi: Strict root comparison equality endpoint 2 ni chiqaradi.

⚠️ Source 8118 x=2 ni noto‘g‘ri yopgan.

ortacha $(x-1)\sqrt{-x^2+x+6}\ge0$ tengsizlikni yeching.

💡 Maslahat: Radical domain va radical-zero endpointsni ajrating.

  1. Radikand -(x-3)(x+2)≥0 → domain [-2,3].
  2. Interior (-2,3) da radical positive, shuning uchun sign x-1 bilan bir xil.
  3. Interior solution x≥1 → [1,3).
  4. Radical zero endpoints x=-2 va x=3 productni 0 qiladi va ≥0 ga kiradi.
  5. Natija {-2}∪[1,3].

✅ Javob: $\{-2\}\cup[1,3]$

Nega bu usul ishlaydi: Product sign theorem radical-zero boundaryni algebraic sign intervaliga qo‘shadi.

⚠️ Source 8121 bilan mos.

murakkab $(x^2+2x-8)\sqrt{x^2+x-2}\le0$ tengsizlikni yeching.

💡 Maslahat: Radicand domain va polynomial signini kesishiring.

  1. Radikand (x+2)(x-1)≥0 → x≤-2 yoki x≥1.
  2. Radical positive interiorlarda product signini P=(x+4)(x-2) belgilaydi.
  3. P≤0 uchun -4≤x≤2.
  4. Domain bilan kesishma [-4,-2]∪[1,2].
  5. x=-2 va x=1 radical-zero bo‘lib inclusionni saqlaydi.

✅ Javob: $[-4,-2]\cup[1,2]$

Nega bu usul ishlaydi: Nonnegative radical factor va polynomial sign chart.

Muqobil usul: Full critical-point sign chart ham ishlaydi.

⚠️ Source 8125 bilan mos.

ortacha $\frac{6-x}{\sqrt{x^2-8x+7}}\ge0$ tengsizlikni yeching.

💡 Maslahat: Denominator root sabab radikand strict positive.

  1. Denominator uchun x²-8x+7=(x-1)(x-7)>0.
  2. Domain x<1 yoki x>7.
  3. Denominator positive, shuning uchun numerator 6-x≥0 →x≤6.
  4. Kesishma x<1.

✅ Javob: $(-\infty,1)$

Nega bu usul ishlaydi: Positive radical denominator inequality signini o‘zgartirmaydi, lekin endpoints 1 va7 poles.

Muqobil usul: Sign chart orqali.

⚠️ Source 8126 x=1 ni noto‘g‘ri kiritgan.

oson $\frac{(x-2)(x-4)}{\sqrt{x^2+x+1}}<0$ tengsizlikni yeching.

💡 Maslahat: Denominator har doim positive ekanini tekshiring.

  1. x²+x+1 diskriminanti -3<0 va leading coefficient positive, demak radikand >0 barcha real x da.
  2. Denominator strictly positive.
  3. Shuning uchun (x-2)(x-4)<0.
  4. Natija 2<x<4.

✅ Javob: $(2,4)$

Nega bu usul ishlaydi: Positive denominator signni numeratorga qoldiradi.

⚠️ Source 8127 endpointlarni noto‘g‘ri yopgan.

murakkab $|x-5|\sqrt{x^2-49}(x+3)\le0$ tengsizlikni yeching.

💡 Maslahat: Domain ikki tashqi intervaldan iborat.

  1. Domain x²-49≥0 → x≤-7 yoki x≥7.
  2. |x-5|≥0 va radical≥0.
  3. x<-7 da qolgan factor x+3<0, product negative; x=-7 da radical zero.
  4. x>7 da x+3>0 va boshqa factors positive, product positive; x=7 da radical zero.
  5. Natija (-∞,-7]∪{7}.

✅ Javob: $(-\infty,-7]\cup\{7\}$

Nega bu usul ishlaydi: Nonnegative factors signni x+3 ga qoldiradi, zero factors equality endpointlar beradi.

⚠️ Source 8128 `[7,8]∪{-7}` deb bergan; noto‘g‘ri.

murakkab $\frac{|x+7|\sqrt{x^2-25}(x-8)}{x+9}<0$ tengsizlikni yeching.

💡 Maslahat: Domain va barcha zero/pole critical pointlarni joylashtiring.

  1. Domain x≤-5 yoki x≥5; pole x=-9.
  2. Strict inequalityda zeros x=-7, -5, 5, 8 solution emas.
  3. |x+7| va sqrt factor interiorlarda positive; sign (x-8)/(x+9) bilan boshqariladi, x=-7 da alohida zero bor.
  4. Negative intervals (-9,-7), (-7,-5), (5,8).

✅ Javob: $(-9,-7)\cup(-7,-5)\cup(5,8)$

Nega bu usul ishlaydi: Absolute value va radical nonnegative factors; rational sign chart strict endpointsni chiqaradi.

⚠️ Source 8129 x=5 ni noto‘g‘ri yopgan.

ortacha $\frac{x-\sqrt{x}-2}{x-\sqrt{x}-6}>0$ tengsizlikni yeching.

💡 Maslahat: t=sqrt(x)≥0.

  1. t=√x≥0, x=t².
  2. Ratio (t²-t-2)/(t²-t-6)=((t-2)(t+1))/((t-3)(t+2)).
  3. t≥0 da t+1,t+2 positive, shuning uchun (t-2)/(t-3)>0.
  4. Solution 0≤t<2 yoki t>3.
  5. x=t² → [0,4)∪(9,∞).

✅ Javob: $[0,4)\cup(9,\infty)$

Nega bu usul ishlaydi: sqrt substitution rational inequalityni oddiy sign chartga aylantiradi.

⚠️ Source 8130 bilan mos.

murakkab $\sqrt{x}-3\le\frac{2}{\sqrt{x}-2}$ tengsizlikni yeching.

💡 Maslahat: t=sqrt(x)≥0 va t≠2.

  1. t=√x≥0, t≠2.
  2. Bir tomonga: (t-3)-2/(t-2)≤0.
  3. Common denominator: ((t-1)(t-4))/(t-2)≤0.
  4. t≥0 sign chart → [0,1]∪(2,4].
  5. x=t² → [0,1]∪(4,16].

✅ Javob: $[0,1]\cup(4,16]$

Nega bu usul ishlaydi: Substitution pole t=2 ni ochiq saqlaydi; nonstrict zeros 1 va4 kiradi.

⚠️ Source 8131 x=4 ni noto‘g‘ri kiritgan; original denominator 0.

ortacha $2\sqrt{x}-3\sqrt[4]{x}+1\ge0$ tengsizlikni yeching.

💡 Maslahat: u=fourthroot(x)≥0.

  1. u=x^(1/4)≥0, sqrt(x)=u².
  2. 2u²-3u+1≥0.
  3. (2u-1)(u-1)≥0.
  4. u∈[0,1/2]∪[1,∞).
  5. x=u⁴ → [0,1/16]∪[1,∞).

✅ Javob: $[0,\frac1{16}]\cup[1,\infty)$

Nega bu usul ishlaydi: Fourth-root substitution mixed radical powersni quadraticga tushiradi.

⚠️ Source 8132 bilan mos.

ortacha $\sqrt{x}-5\sqrt[4]{x}+6\ge0$ tengsizlikni yeching.

💡 Maslahat: u=x^(1/4).

  1. u≥0 va sqrt(x)=u².
  2. u²-5u+6≥0 →(u-2)(u-3)≥0.
  3. u≤2 yoki u≥3, u≥0 bilan [0,2]∪[3,∞).
  4. x=u⁴ → [0,16]∪[81,∞).

✅ Javob: $[0,16]\cup[81,\infty)$

Nega bu usul ishlaydi: Quadratic endpoints u=2,3 nonstrict sabab kiradi.

⚠️ Source 8133 marked A x=16 ni chiqarib yuborgan; B canonical.

ortacha $\sqrt{x}-12\sqrt[4]{x}+27\le0$ tengsizlikni yeching.

💡 Maslahat: u=x^(1/4).

  1. u≥0.
  2. u²-12u+27≤0.
  3. (u-3)(u-9)≤0 →3≤u≤9.
  4. x=u⁴ →81≤x≤6561.

✅ Javob: $[3^4,9^4]=[81,6561]$

Nega bu usul ishlaydi: Fourth-root substitution exact endpoint powersni beradi.

⚠️ Source 8136 `[3^6,9^6]` deb belgilagan; noto‘g‘ri.

oson $2\sqrt{x-1}-\sqrt{x-1}\ge1$ tengsizlikni yeching.

💡 Maslahat: Like radical termsni birlashtiring.

  1. LHS sqrt(x-1) ga teng.
  2. Domain x≥1.
  3. sqrt(x-1)≥1 →x-1≥1.
  4. x≥2.

✅ Javob: $[2,\infty)$

Nega bu usul ishlaydi: Soddalashtirishdan keyin positive-threshold master rule.

Muqobil usul: Direct square.

⚠️ Source 8138 faqat {2} ni belgilagan; canonical x≥2.

murakkab $\sqrt{\frac{x-1}{x+1}}-\sqrt{\frac{x+1}{x-1}}<\frac32$ tengsizlikni yeching.

💡 Maslahat: t=sqrt((x-1)/(x+1))>0.

  1. Original domain ratio positive va x≠±1: x<-1 yoki x>1.
  2. t>0 va reciprocal root=1/t.
  3. t-1/t<3/2 →2t²-3t-2<0.
  4. Positive t uchun 0<t<2.
  5. t²=(x-1)/(x+1)<4.
  6. Domain bilan sign solve natijasi (-∞,-5/3)∪(1,∞).

✅ Javob: $(-\infty,-\frac53)\cup(1,\infty)$

Nega bu usul ishlaydi: Reciprocal-root substitution expressionni quadratic t inequalityga tushiradi.

⚠️ Source 8139 -5/3 endpointini noto‘g‘ri yopgan.

ortacha $\sqrt{x+6}>x$ tengsizlikni yeching.

💡 Maslahat: RHS x ishorasiga qarab case split qiling.

  1. Domain x≥-6.
  2. Agar -6≤x<0 bo‘lsa RHS negative, LHS nonnegative; inequality avtomatik rost.
  3. Agar x≥0 bo‘lsa safe square: x+6>x².
  4. x²-x-6<0 → -2<x<3; x≥0 bilan [0,3).
  5. Union [-6,3).

✅ Javob: $[-6,3)$

Nega bu usul ishlaydi: Variable RHS lower-strict master rule negative RHS regionni avtomatik oladi.

⚠️ Source 8140 bilan mos.

murakkab $\sqrt{x+4}>2x+2$ tengsizlikni yeching.

💡 Maslahat: RHS 2x+2 ning ishorasini ajrating.

  1. Domain x≥-4.
  2. 2x+2<0 →x<-1; domain bilan [-4,-1) avtomatik solution.
  3. x≥-1 regionda square: x+4>(2x+2)².
  4. 4x²+7x<0 → -7/4<x<0; x≥-1 bilan [-1,0).
  5. Union [-4,0).

✅ Javob: $[-4,0)$

Nega bu usul ishlaydi: RHS sign split squaringni faqat qonuniy regionda ishlatadi.

⚠️ Source 8141 x=0 ni noto‘g‘ri kiritgan.

murakkab $0<x+\sqrt{3-2x}$ tengsizlikni yeching.

💡 Maslahat: sqrt(3-2x)>-x ko‘rinishiga keltiring.

  1. Domain x≤3/2.
  2. Agar x≥0 bo‘lsa RHS -x≤0 va root≥0; inequality avtomatik, [0,3/2].
  3. Agar x<0 bo‘lsa -x>0 va safe square: 3-2x>x².
  4. x²+2x-3<0 → -3<x<1; x<0 bilan (-3,0).
  5. Union (-3,3/2].

✅ Javob: $(-3,\frac32]$

Nega bu usul ishlaydi: Variable RHS sign split upper/lower comparisonni exact qiladi.

⚠️ Source 8143 bilan mos.

ortacha $x<\sqrt{x+12}$ tengsizlikni yeching.

💡 Maslahat: x<0 branch automatic.

  1. Domain x≥-12.
  2. Agar x<0 bo‘lsa LHS negative, root nonnegative: [-12,0) solution.
  3. Agar x≥0 bo‘lsa square: x²<x+12.
  4. x²-x-12<0 → -3<x<4; x≥0 bilan [0,4).
  5. Union [-12,4).

✅ Javob: $[-12,4)$

Nega bu usul ishlaydi: sqrt(A)>B formula B=x bilan ishlaydi.

⚠️ Source 8145 x=4 ni noto‘g‘ri yopgan.

ortacha $f(x)=\sqrt{7-2\sqrt{x-1}}$ funksiyaning real domainini toping.

💡 Maslahat: Inner va outer root shartlarini ketma-ket yozing.

  1. Inner root: x-1≥0 →x≥1.
  2. Outer radicand: 7-2sqrt(x-1)≥0.
  3. sqrt(x-1)≤7/2.
  4. Square: x-1≤49/4 →x≤53/4.
  5. Domain [1,53/4].

✅ Javob: $[1,\frac{53}{4}]$

Nega bu usul ishlaydi: Nested radical domain ichki va tashqi constraints kesishmasidir.

⚠️ Source 8146 bilan mos.

ortacha $\sqrt{11}-\sqrt{10}>\sqrt{12}-\sqrt{11}$ rostmi?

💡 Maslahat: Conjugate orqali ketma-ket square-root farqlarini taqqoslang.

  1. d(t)=sqrt(t+1)-sqrt(t)=1/(sqrt(t+1)+sqrt(t)).
  2. Denominator t ortishi bilan ortadi, demak d(t) kamayadi.
  3. Chap tomon d(10), o‘ng tomon d(11).
  4. d(10)>d(11), tengsizlik rost.

✅ Javob: Tengsizlik rost

Nega bu usul ishlaydi: Conjugate difference positive va decreasing sequence ekanini ko‘rsatadi.

⚠️ Source 8149 variable yo‘q bo‘lsa ham `∅` ni marked qilgan; matematik proposition aslida rost.

murakkab $\sqrt{x}-\sqrt{x-1}>\sqrt{x-2}-\sqrt{x-3}$ tengsizlikni yeching.

💡 Maslahat: Root-difference function kamayishini ishlating.

  1. Domain x≥3.
  2. d(t)=sqrt(t)-sqrt(t-1)=1/(sqrt(t)+sqrt(t-1)).
  3. d(t) t ortishi bilan qat’iy kamayadi.
  4. Chap tomon d(x), o‘ng tomon d(x-2).
  5. x>x-2 bo‘lgani uchun d(x)<d(x-2), so‘ralgan > hech qachon bajarilmaydi.

✅ Javob: $\varnothing$

Nega bu usul ishlaydi: Conjugate monotonicity repeated squaringni butunlay chetlab o‘tadi.

⚠️ Source 8151 variantlari hatto domain x≥3 ga ham mos emas.

murakkab $\sqrt{x-3}+\sqrt{2x-1}\le\sqrt{2x+2}$ tengsizlikni yeching.

💡 Maslahat: Domain x≥3; square qilganda qolgan root RHSini nonnegative saqlang.

  1. Domain x≥3.
  2. Square: 3x-4+2sqrt((x-3)(2x-1))≤2x+2.
  3. 2sqrt((x-3)(2x-1))≤6-x, demak x≤6 zarur.
  4. Ikkala tomon nonnegative region [3,6] da yana square: 4(x-3)(2x-1)≤(6-x)².
  5. Bu 7x²-16x-24≤0.
  6. Roots (8±2sqrt58)/7; [3,6] bilan kesishma [3,(8+2sqrt58)/7].

✅ Javob: $[3,\frac{8+2\sqrt{58}}7]$

Nega bu usul ishlaydi: Repeated squaringda har yangi RHS sign conditioni saqlanadi.

⚠️ Source 8152 variantlari domain tashqarisida; canonical interval shu.

ortacha $\sqrt{x-3}+\sqrt{3x-7}\ge\sqrt{4x-16}$ tengsizlikni yeching.

💡 Maslahat: Common domain va root-sum lower bound.

  1. Common domain x≥4.
  2. A=x-3, B=3x-7; A+B=4x-10.
  3. Root-sum theorem: LHS≥sqrt(4x-10).
  4. x≥4 da 4x-10>4x-16, demak sqrt(4x-10)>sqrt(4x-16).
  5. Shuning uchun barcha x≥4 solution.

✅ Javob: $[4,\infty)$

Nega bu usul ishlaydi: Structural bound direct comparisondan kuchliroq lower bound beradi.

⚠️ Source 8153 `[3,∞)` deb belgilagan; x=3 da RHS radikandi manfiy.

murakkab $(x^2-16)\sqrt{\frac{x^2-x-6}{3x^2-16x+16}}\le0$ tengsizlikni yeching.

💡 Maslahat: Radicandni factorlab domain sign chartini toping.

  1. Radicand=(x-3)(x+2)/((3x-4)(x-4)).
  2. Domain sign chart: (-∞,-2]∪(4/3,3]∪(4,∞). Poles x=4/3,4 chiqariladi.
  3. Radical positive domain interiorida product signini x²-16 belgilaydi; x²-16≤0 uchun -4≤x≤4.
  4. Domain bilan kesishma [-4,-2]∪(4/3,3].
  5. Radical-zero x=-2,3 nonstrict inequalityga kiradi.

✅ Javob: $[-4,-2]\cup(\frac43,3]$

Nega bu usul ishlaydi: Radicand domain + nonnegative radical factor + polynomial sign intersection.

Muqobil usul: Bitta combined critical-point sign chart qurish mumkin.

⚠️ Source 8155 va 8156 aynan bir prompt uchun ikki xil javob beradi; ikkalasi ham canonical setga mos emas.

murakkab $\frac{\sqrt{5x-1}}{x-2}\ge\frac{\sqrt{5x-1}}{x+1}$ tengsizlikni yeching.

💡 Maslahat: Radical zero branchni cancellationdan oldin ajrating.

  1. Domain x≥1/5, x≠2; x+1>0 bu domainda.
  2. x=1/5 da common numerator 0, inequality 0≥0 rost.
  3. x>1/5 da sqrt(5x-1)>0, cancel qilamiz.
  4. 1/(x-2)≥1/(x+1) →3/((x-2)(x+1))≥0.
  5. x+1>0, demak x>2.
  6. Natija {1/5}∪(2,∞).

✅ Javob: $\{\frac15\}\cup(2,\infty)$

Nega bu usul ishlaydi: Zero numerator branch cancellationda yo‘qolmasligi kerak; positive branchda safe cancellation.

Muqobil usul: Common denominator sign chart.

⚠️ Source 8158 x=2 poleni noto‘g‘ri kiritgan.

murakkab $x+\frac{x}{\sqrt{x^2-1}}>\frac{35}{12}$ tengsizlikni yeching.

💡 Maslahat: Domain |x|>1; negative branchni sign bilan chiqarib tashlang.

  1. Domain |x|>1.
  2. x<-1 da x<0 va x/sqrt(x²-1)<0, shuning uchun LHS<0<35/12; bu branchda solution yo‘q.
  3. x>1 da s=sqrt(x²-1)>0 deb oling. Inequality 12x(s+1)>35s, ya’ni 12x>(35-12x)s.
  4. Agar x≥35/12 bo‘lsa RHS≤0, LHS>0; inequality avtomatik rost.
  5. 1<x<35/12 da ikkala tomon positive; square qilish ekvivalent: 144x²>(35-12x)²(x²-1).
  6. Difference factorlanadi: 144x²-(35-12x)²(x²-1)=-(3x-5)(4x-5)(12x²-35x-49).
  7. 1<x<35/12 da 12x²-35x-49<0, demak condition (3x-5)(4x-5)>0 ga tushadi.
  8. Bu 1<x<5/4 yoki 5/3<x<35/12 ni beradi; x≥35/12 automatic branch bilan birlashtirib (1,5/4)∪(5/3,∞).

✅ Javob: $(1,\frac54)\cup(\frac53,\infty)$

Nega bu usul ishlaydi: Positive denominator orqali transform qilingan inequalityda faqat positive-positive branch square qilinadi; exact factorization critical points 5/4 va 5/3 ni beradi.

Muqobil usul: Calculus bilan f(x)=x+x/sqrt(x²-1) ning minimum strukturasini tahlil qilish mumkin, lekin exact algebra source QA uchun kuchliroq.

⚠️ Source 8172 x=5/4 endpointini noto‘g‘ri yopgan.

murakkab $\frac1{\sqrt{x+2\sqrt{x-1}}}+\frac1{\sqrt{x-2\sqrt{x-1}}}>2$ tengsizlikni yeching.

💡 Maslahat: t=sqrt(x-1)≥0; radikandlar perfect squares.

  1. t=√(x-1)≥0, x=t²+1.
  2. x±2√(x-1)=t²+1±2t=(t±1)².
  3. Denominators |t+1|=t+1 va |t-1|; t=1 (x=2) pole.
  4. 0≤t<1 da expression 1/(t+1)+1/(1-t)=2/(1-t²)>2 iff t>0 →0<t<1.
  5. t>1 da expression 1/(t+1)+1/(t-1)=2t/(t²-1)>2 → t²-t-1<0.
  6. Thus 1<t<(1+sqrt5)/2.
  7. Back-substitute x=t²+1: (1,2)∪(2,(5+sqrt5)/2).

✅ Javob: $(1,2)\cup(2,\frac{5+\sqrt5}{2})$

Nega bu usul ishlaydi: Nested perfect squares absolute values va pole t=1 ni aniq ochadi.

Muqobil usul: Har branchda common denominator ishlatish mumkin.

⚠️ Source 8173 x=1 va upper equality endpointini noto‘g‘ri kiritgan.

murakkab $\sqrt{\frac1{x^2}}-\frac34+\frac1x+\frac12<0$ tengsizlikni yeching.

💡 Maslahat: sqrt(1/x²)=1/|x|.

  1. Domain x≠0.
  2. Expression 1/|x|+1/x-1/4<0.
  3. x<0 da 1/|x|=-1/x, birinchi ikki had bekor bo‘ladi; -1/4<0, barcha x<0 solution.
  4. x>0 da 1/|x|=1/x, demak 2/x<1/4.
  5. x>0 bilan x>8.

✅ Javob: $(-\infty,0)\cup(8,\infty)$

Nega bu usul ishlaydi: Absolute value case split rational inequalityni ikki sodda branchga ajratadi.

⚠️ Source 8175 variantlari canonical setga umuman mos emas.

ortacha $\sqrt{16-x^2}+\sqrt{9-x^2}\ge7$ tengsizlikni yeching.

💡 Maslahat: Har rootning alohida maksimumini ishlating.

  1. Common domain |x|≤3.
  2. sqrt(16-x²)≤4 va sqrt(9-x²)≤3.
  3. Shuning uchun LHS≤7.
  4. ≥7 bo‘lishi uchun ikkala upper bound bir vaqtda equality bo‘lishi kerak.
  5. Bu faqat x=0 da sodir bo‘ladi; LHS=4+3=7.

✅ Javob: $\{0\}$

Nega bu usul ishlaydi: Extremal upper bounds equality condition bilan exact solution beradi.

Muqobil usul: Function symmetry va monotonicity |x| bo‘yicha ham ishlaydi.

⚠️ Source 8166 bilan mos.

ortacha $\sqrt{x-4}+\sqrt{x+4}\ge\sqrt{2x}$ tengsizlikni yeching.

💡 Maslahat: Common domain x≥4; square structural comparison.

  1. Common domain x≥4.
  2. Ikkala tomon nonnegative.
  3. LHS²=2x+2sqrt(x²-16)≥2x=RHS².
  4. Shuning uchun barcha x≥4 solution.

✅ Javob: $[4,\infty)$

Nega bu usul ishlaydi: Cross term nonnegative bo‘lgani uchun root-sum structural bound ishlaydi.

⚠️ Source 8154 variantlari 3/2 atrofida va original domain bilan mos emas.

Umumiy xatolar

❌ Domainni yozmasdan tengsizlikni algebraik yechish.

Radikal real bo‘lmagan intervallar soxta solutionga kirishi mumkin.

✅ Birinchi qadamda barcha radikand va denominator shartlarini yozing.

sqrt(3x-10)>sqrt(6-x) uchun 10/3≤x≤6.

❌ sqrt(A)<negative constant ni square qilish.

LHS nonnegative; aslida bunday tengsizlik darhol yechimsiz.

✅ Principal-root signidan foydalaning.

sqrt(x)<-14 → ∅.

❌ sqrt(A)>negative constant ni square qilish.

Square qilganda direction/solution region noto‘g‘ri qisqarishi mumkin; aslida butun domain solution.

✅ Negative RHS bo‘lsa domainning o‘zini oling.

sqrt(x-8)>-6 → x≥8.

❌ sqrt(A)≤0 ni A≤0 deb yozish.

Root real bo‘lishi uchun A≥0 ham zarur; ikkalasi A=0 beradi.

✅ sqrt(A)≤0 iff A=0.

sqrt(3x+5)≤0 → x=-5/3.

❌ Strict inequality endpointini yopish.

Equality boundary < yoki > ni qanoatlantirmaydi.

✅ Algebraic equality rootsni open endpoint qiling.

sqrt(x)>2 → (4,∞), 4 emas.

❌ Radical denominator radikandi uchun ≥0 ishlatish.

Radikand 0 bo‘lsa denominator 0.

✅ Denominator sqrt(Q) uchun Q>0.

(6-x)/sqrt((x-1)(x-7)) da 1 va7 chiqariladi.

❌ Rational radicandda denominator zero nuqtani inclusion qilish.

P/Q radikand xuddi kasr kabi polega ega.

✅ Q≠0 ni sign chartga explicit kiriting.

sqrt((x-2)/(x-1)) da x=1 forbidden.

❌ sqrt(A)<B(x) da B ishorasini tekshirmay square qilish.

B≤0 bo‘lsa nonnegative LHS undan kichik bo‘la olmaydi.

✅ B>0 conditionni yozib, faqat keyin square qiling.

x<sqrt(x+12) ni x signi bo‘yicha ajratish.

❌ sqrt(A)>B(x) da negative B regionni yo‘qotish.

Negative RHS regionda inequality avtomatik rost bo‘lishi mumkin.

✅ B<0 regionni domain bilan darhol solutionga qo‘shing.

sqrt(x+6)>x da [-6,0) automatic.

❌ sqrt(A)?sqrt(B) da faqat A?B ni yechib domainni unutish.

Har ikkala radikand nonnegative bo‘lishi kerak.

✅ Common domainni oldin toping.

sqrt(3x-10)>sqrt(6-x).

❌ Radikal productda radical factorni doim positive deb olish.

Radical domain boundaryda zero bo‘lishi mumkin va product equalityni qanoatlantiradi.

✅ Q>0 interior va Q=0 endpointsni alohida tekshiring.

(x-1)sqrt(-x²+x+6)≥0 da x=-2,3.

❌ Radikal quotientda denominator signini noma’lum deb interval almashtirish.

sqrt(Q)>0 domain ichida denominator strictly positive.

✅ Signni numerator orqali o‘qing.

(x-2)(x-4)/sqrt(x²+x+1)<0.

❌ t=sqrt(x) substitutiondan keyin t<0 ildizlarni qabul qilish.

Principal root sabab t≥0.

✅ Yangi o‘zgaruvchi constraintni sign chartga qo‘shing.

t=√x.

❌ u=x^(1/4) substitutionda x=u² deb qaytarish.

Fourth root uchun x=u⁴.

✅ sqrt(x)=u², x=u⁴ ni birga yozing.

sqrt(x)-5 fourthroot(x)+6≥0.

❌ Cancellationdan keyin original pole nuqtani qayta kiritish.

Equivalent simplified formula faqat original domain ichida.

✅ Original denominator restrictionsni yakunda saqlang.

sqrt((x-7)/(3x-21)) da x=7 forbidden.

❌ Repeated squaringni har masalaga default qilish.

Yuqori darajali ifoda, extraneous region va endpoint xatolari ko‘payadi.

✅ Monotonicity, substitution, conjugate yoki boundsni avval qidiring.

Root-difference 8151 conjugate bilan darhol ∅.

❌ Nested radicalda faqat inner domainni olish.

Outer radicand ham nonnegative bo‘lishi kerak.

✅ Barcha nested constraintsni kesishiring.

sqrt(7-2sqrt(x-1)) → [1,53/4].

❌ Approximate decimal bilan strict endpoint inclusion qaror qilish.

Yaqinlik equality/non-equalityni isbotlamaydi.

✅ Exact radicals va factorizationdan foydalaning.

(8+2sqrt58)/7 endpointi.

❌ Marked source optionni canonical deb olish.

Bankda poles, endpoints va hatto duplicate contradictions mavjud.

✅ Independent exact algebra bilan tekshiring.

8155 va8156 bir prompt, ikki xil javob.

❌ All-domain yoki no-solution holatda uzoq algebra qilish.

Simple lower/upper bound masalani bir qadamda hal qilishi mumkin.

✅ Extremal boundni avval tekshiring.

sqrt(4x²-12x+41)≥sqrt32>2.

Noto'g'ri tasavvurlar

Tengsizlikda square qilish tenglamadagidek doim xavfsiz.

Faqat taqqoslanayotgan ikki tomon nonnegative bo‘lganda square order-equivalent.

Square root manfiy qiymat ham olishi mumkin.

Principal square root har doim nonnegative.

sqrt(A)≤0 ning yechimi A≤0.

Real-domain bilan birga faqat A=0 qoladi.

≥ belgisi bo‘lsa denominator pole endpointini qo‘shish mumkin.

Undefined point hech qanday inequality solutioni bo‘la olmaydi.

sqrt(A)>B da doim A>B².

B<0 regionda original domain avtomatik solution; square formula faqat B≥0 branchda.

Ikki square rootni solishtirish uchun domain kerak emas.

A va B ikkalasi ham nonnegative bo‘lishi kerak.

Radikal factor product ishorasini murakkablashtiradi.

Principal radical domain interiorida positive, boundaryda zero; sign tahlili aksincha soddalashadi.

Substitution faqat tenglamalarda ishlaydi.

t=√x va u=x^(1/4) inequalitylarda ham sign chartni soddalashtiradi, faqat t,u≥0 saqlanadi.

Strict inequalityda source endpoint bracketi ishonchli.

Endpoint original expressionda exact equality/pole bilan qayta tekshiriladi.

Ko‘p radikalli tengsizlik faqat ketma-ket square bilan yechiladi.

Root-sum bounds, conjugate monotonicity va extremal estimates ko‘pincha qisqaroq va xavfsizroq.

Amaliy qo'llanilishi

Funksiya va grafiklar

Radikal funksiyaning domaini, musbat/manfiy regioni va grafiklar kesishmasini inequality sifatida tahlil qilish.

Geometriya

Masofa va radius shartlari “uzunlik kamida/ko‘pi bilan” ko‘rinishida radical inequalities beradi.

Fizika

Tezlik, energiya va periodning real hamda physical sign constraintslarini inequality bilan tekshirish.

Engineering

Tolerance, stress va design limitslarda square-root modelning admissible parameter intervalini topish.

Optimization

Radical expressionning minimum/maximum boundlari feasible region va equality conditionni beradi.

Numerical analysis

Approximate root/interval natijasini exact domain va endpoint bilan validatsiya qilish.

Computer algebra

CAS transformationsda squaring, branch va principal-root assumptionsni tekshirish false inequality equivalenceni oldini oladi.

Test-bank QA

Strict endpoints, denominator poles, domain intersections va bounds OCR/marked-answer xatolarini tez aniqlaydi.

Irratsional tengsizlik: sign-first yechim xaritasi

Irratsional tengsizlik — DOMAIN → SIGN → SAFE TRANSFORM1. ORIGINAL DOMAINeven radicand ≥0 • radical denominator >0 • poles outcommon domain before comparing rootsROOT vs CONSTANTRHS <0 → shortcutRHS ≥0 → safe squarestrict ↔ open equalitynonstrict ↔ boundary checkSIGN / SUBSTITUTIONP√Q → radical 0/+P/√Q → Q>0t=√x≥0u=x^(1/4)≥0STRUCTURE / BOUNDS√A ? √B → A ? Bconjugate differencesroot-sum boundsmin/max before squaringFINAL INTERVAL QAintersect original domain • remove poles • test equality boundariesstrict/open vs nonstrict/closed • marked option ≠ proof

Domain va RHS signidan boshlanib constant, sign/substitution yoki structure/bounds yo‘li tanlanadi; yakunda interval endpointlari va poles original expressionda tekshiriladi.

Xulosa

Cheat sheet: sqrt(A) real iff A>=0; 1/sqrt(A) defined iff A>0. sqrt(A)<0 — yechim yo‘q; sqrt(A)<=0 iff A=0; sqrt(A)>0 iff A>0; sqrt(A)>=0 iff A>=0. c>0 bo‘lsa sqrt(A)<c iff 0<=A<c² va sqrt(A)<=c iff 0<=A<=c². c<0 bo‘lsa sqrt(A)>c va sqrt(A)>=c original domainda avtomatik. Common domain ostida sqrt(A)?sqrt(B) iff A?B. Har strict inequalityda equality endpointlari chiqariladi; denominator poles hech qachon kiritilmaydi.

Keyingi bosqichlarda shu domain-first va sign-first fikrlash funksiyalar, ko‘rsatkichli/logarifmik tengsizliklar hamda parametrli masalalarda davom etadi. Ayniqsa monotonic function orqali inequality transform qilish va endpoint QA umumiy metod bo‘lib qoladi.

Bog'liq mavzular

Oldin bilishingiz kerak: Ratsional ko'rsatkichli daraja va uning xossalari, Irratsional tenglamalar, Ko'phadlar va ular ustida amallar

Bog'liq mavzular: Modulli ifodalar va tenglamalar, Funksiya

Keyingi mavzular: Ko'rsatkichli tengsizliklar, Ko'rsatkichli tenglama, Logarifmlar: hisoblashga doir masalalar

Manbalar

Ro'yxatdan o'tib, mashq qilishni boshlang