MathTest.uz
Algebra

Ratsional ko'rsatkichli daraja va uning xossalari

murakkab 190 daqiqa ratsional ko‘rsatkichfractional exponentradikalildizdaraja xossalariprincipal rootabsolute valuemanfiy ko‘rsatkichrationalizationconjugatenested radicaldomain

Nima uchun muhim?

Ratsional ko‘rsatkich daraja va ildiz yozuvlarini bitta algebraik tizimga birlashtiradi. Bu mavzu irratsional tenglama va tengsizliklar, ko‘rsatkichli funksiyalar, logarifmlar, formulalarni soddalashtirish va algebraik transformatsiyalarning asosidir. Asosiy xavf — daraja qonunlarini domain va juft/toq ildiz shartlarisiz mexanik qo‘llash.

O'quv maqsadlari

  • Butun ko‘rsatkichli daraja qonunlarini ratsional ko‘rsatkichlarga bog‘lash
  • a^(1/n) ni n-darajali ildiz sifatida tushuntirish
  • a^(m/n) ni radical formaga va aksincha o‘tkazish
  • Kasr ko‘rsatkichni eng sodda ko‘rinishda domain bilan talqin qilish
  • Manfiy ratsional ko‘rsatkichni reciprocal orqali soddalashtirish
  • Bir xil asosli darajalarni ko‘paytirish va bo‘lish
  • Darajaning darajasi qoidasini domain sharti bilan ishlatish
  • Ko‘paytma va kasrni ratsional darajaga ko‘tarish
  • Juft indeksli ildiz uchun radikand nonnegative shartini aniqlash
  • Toq indeksli ildizda manfiy radikand mumkinligini tushuntirish
  • sqrt[n](a^n) da juft n uchun |a| paydo bo‘lishini qo‘llash
  • Radikallarni perfect-power factor orqali soddalashtirish
  • Radikal ko‘paytma va kasr xossalarini real-domain shartlari bilan ishlatish
  • Turli indeksli musbat radikallarni umumiy darajaga ko‘tarib taqqoslash
  • Bir hadli maxrajdagi radikalni ratsionallashtirish
  • Ikki hadli kvadrat ildizli maxrajni conjugate bilan ratsionallashtirish
  • Kub ildizli strukturada sum/difference-of-cubes multiplierdan foydalanish
  • Nested radicalsni rational exponent orqali bitta darajaga keltirish
  • Radical expressionlarda common factor va like radicalsni ajratish
  • Infinite radical uchun fixed-point tenglama va admissible rootni tanlash
  • Source-bankdagi real-domain va extraction xatolarini mustaqil aniqlash
  • Keyingi irratsional tenglamalar uchun domain-first fikrlashni tayyorlash
Ratsional ko‘rsatkich — ildizning daraja yozuvidagi shakli. a>0 uchun a^(m/n)=sqrt[n](a^m)=(sqrt[n](a))^m. Lekin real sonlarda n juft bo‘lsa ildiz osti manfiy bo‘la olmaydi; sqrt(a²)=|a|, a emas. Shuning uchun daraja qonunlarining “ko‘rinishi” oddiy bo‘lsa ham, ularni qo‘llashdan oldin base, denominator parity va denominator nol bo‘lmasligi kabi shartlar tekshiriladi. Radikal va daraja yozuvlari orasida erkin o‘tish ko‘p ifodani bir xil asosli darajalar masalasiga aylantiradi. Masalan sqrt(x)=x^(1/2), cubert(x²)=x^(2/3), x^p·x^q=x^(p+q). Soddalashtirishda perfect-power factorlarni ajratish, manfiy ko‘rsatkichni reciprocalga aylantirish va juft ildizdan chiqqan absolute value’ni saqlash muhim. Source-bankda ko‘plab extraction xatolari bor: ayrim savollarda ko‘rsatkich yoki ildiz indeksi yo‘qolgan, ayrim marked javoblar algebraik ifodaga mos emas, ayrim juft ildizlarda negative radicand real sohada aniqlanmagan. Canonical dars marked optionga emas, real-domain qoidalari va algebraik tekshiruvga tayanadi.

Ta'riflar

Ratsional ko‘rsatkichli daraja · Rational exponent

$a^{m/n}$ — kasr ko‘rsatkichli daraja; real talqin domain shartlariga bog‘liq.

Ildiz va darajani bitta yozuvga birlashtiradi.

Misol: $8^{2/3}=(\sqrt[3]{8})^2=4$.

Bu emas: $(-4)^{1/2}$ real son emas.

💡 Kasr m/n odatda qisqartirilgan ko‘rinishda talqin qilinadi.

n-darajali ildiz · nth root

$\sqrt[n]{a}$ — n-darajaga ko‘tarilganda a beradigan principal real ildiz.

Radikal belgisi ostidagi son radikand, n esa indeks.

Misol: $\sqrt[3]{-8}=-2$.

Bu emas: $\sqrt[4]{-16}$ real emas.

💡 n≥2.

Kvadrat ildiz · Square root

$\sqrt a$ — a≥0 ning nonnegative kvadrat ildizi.

Principal square root har doim ≥0.

Misol: $\sqrt{25}=5$.

Bu emas: $\sqrt{25}=\pm5$ noto‘g‘ri; ± tenglama yechimida paydo bo‘ladi.

💡 Domain a≥0.

Principal ildiz · Principal root

Juft indeksli real ildizda nonnegative root tanlanadi.

Radikal belgisi bitta principal qiymat beradi.

Misol: $\sqrt{x^2}=|x|$.

Bu emas: $\sqrt{x^2}=x$ barcha x uchun emas.

💡 Absolute value juda muhim.

Radikand · Radicand

Radikal belgisi ichidagi ifoda.

Ildiz olinayotgan obyekt.

Misol: $\sqrt[3]{2x-1}$ da radikand 2x-1.

Bu emas: Butun radical expression radikand emas.

💡 Even-index domain radikand signiga bog‘liq.

Ildiz indeksi · Index

Radikalning n soni; sqrt[n](a) da n.

Qaysi darajali ildiz olinayotganini bildiradi.

Misol: $\sqrt[5]{32}$ da indeks 5.

Bu emas: Oddiy sqrt da indeks 2 yashirin.

💡 n musbat integer, odatda n≥2.

Manfiy ko‘rsatkich · Negative exponent

$a^{-r}=1/a^r$, a≠0.

Minus ishorasi base signini emas, reciprocalni bildiradi.

Misol: $16^{-1/2}=1/4$.

Bu emas: $a^{-2}=-a^2$ noto‘g‘ri.

💡 Base 0 bo‘la olmaydi.

Nol ko‘rsatkich · Zero exponent

$a^0=1$, a≠0.

Har nonzero base ning 0-darajasi 1.

Misol: $7^0=1$.

Bu emas: $0^0=1$ bu kursda qabul qilinmaydi.

💡 a≠0.

Perfect n-th power · To‘liq n-daraja

$b^n$ ko‘rinishidagi factor n-darajali ildizdan tashqariga chiqishi mumkin.

Radikalni soddalashtirish uchun perfect power factor qidiriladi.

Misol: $\sqrt[3]{54}=3\sqrt[3]{2}$.

Bu emas: $\sqrt{12}=\sqrt6\sqrt2$ soddalashtirish emas.

💡 Even n da variable factor absolute value talab qilishi mumkin.

Juft indeksli ildiz · Even-index root

n juft bo‘lsa real sqrt[n](a) uchun a≥0.

Negative radikand real sohada mumkin emas.

Misol: $\sqrt[4]{16}=2$.

Bu emas: $\sqrt[4]{-16}$ real emas.

💡 Complex sonlar alohida mavzu.

Toq indeksli ildiz · Odd-index root

n toq bo‘lsa har real a uchun sqrt[n](a) real.

Manfiy sonning toq ildizi manfiy bo‘ladi.

Misol: $\sqrt[3]{-27}=-3$.

Bu emas: $\sqrt[3]{-27}=3$ noto‘g‘ri.

💡 Odd root signni saqlaydi.

Radikalni soddalashtirish · Simplifying radicals

Perfect powersni radikaldan chiqarib eng sodda equivalent formaga keltirish.

Indeksga mos to‘liq darajalarni ajratamiz.

Misol: $\sqrt{72}=6\sqrt2$.

Bu emas: $\sqrt{72}=\sqrt{36}\sqrt2$ yakuniy soddalashtirilgan ko‘rinish emas.

💡 Domain saqlanadi.

Like radicals · O‘xshash radikallar

Indeksi va soddalashtirilgan radikandi bir xil radikallar.

Faqat shunday radikallar coefficientlar kabi qo‘shiladi.

Misol: $2\sqrt3+5\sqrt3=7\sqrt3$.

Bu emas: $\sqrt2+\sqrt3=\sqrt5$ noto‘g‘ri.

💡 Oldin soddalashtirish foydali.

Maxrajni ratsionallashtirish · Rationalizing denominator

Equivalent kasrda maxrajdan radikalni yo‘qotish.

Numerator va denominatorni mos radical/conjugate bilan ko‘paytiramiz.

Misol: $1/\sqrt5=\sqrt5/5$.

Bu emas: Faqat denominatorni ko‘paytirish mumkin emas.

💡 Original denominator nonzero.

Qo‘shma ifoda · Conjugate

a+b ning conjugate’i a-b va aksincha.

Ko‘paytirilganda middle radical terms yo‘qoladi.

Misol: $(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)=1$.

Bu emas: Bir xil ishorali juftlik conjugate emas.

💡 Difference of squaresga tayanadi.

Domain · Aniqlanish sohasi

Ifoda real va aniqlangan bo‘ladigan variable qiymatlari to‘plami.

Soddalashtirishdan oldin yozilgan restrictionlar keyin ham saqlanadi.

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

Bu emas: $1/\sqrt{x-2}$ uchun x≥2 noto‘g‘ri, denominator 0 bo‘ladi.

💡 Canonical QA gate.

Darajalar qonuni · Exponent law

Bir xil base yoki power structures uchun algebraik qoidalar majmui.

$a^r a^s=a^{r+s}$ kabi.

Misol: $x^{1/2}x^{1/3}=x^{5/6}$ mos domain bilan.

Bu emas: $a^r+b^r=(a+b)^r$ umumiy qonun emas.

💡 Ratsional r,s da real-domain caveat bor.

Umumiy indeks · Common index

Turli radikallarni taqqoslash/ko‘paytirish uchun indekslarni LCM orqali moslashtirish.

Kasr ko‘rsatkichlar uchun common denominator olish bilan bir xil.

Misol: $a^{1/3}$ va $b^{1/4}$ ni 12-daraja orqali solishtirish mumkin.

Bu emas: Indekslarni shunchaki qo‘shish mumkin emas.

💡 Positive quantitiesda comparison xavfsiz.

Nested radical · Ichma-ich ildiz

Radikal ichida yana radical qatnashgan ifoda.

Rational exponentda ko‘rsatkichlar ko‘payadi.

Misol: $\sqrt{\sqrt[3]{x}}=x^{1/6}$, mos domain bilan.

Bu emas: $\sqrt{\sqrt x}=x$ emas.

💡 Even roots domainini tekshiring.

Cheksiz ildiz · Infinite radical

O‘zini takrorlaydigan nested radical limit sifatida talqin qilinadi.

Convergence bo‘lsa fixed-point equation hosil bo‘ladi.

Misol: $x=\sqrt{c x}$ kabi.

Bu emas: Har algebraik root avtomatik admissible emas.

💡 Nonnegative va convergence shartlari tekshiriladi.

Fundamental tushunchalar

Exponent–radical bridge

Ratsional ko‘rsatkichning denominator qismi ildiz indeksini, numerator qismi darajani beradi.

$a^{m/n}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m$

Real talqin uchun root mavjud bo‘lishi kerak.

Fraction reduction matters

Masalan 2/4=1/2; real exponent talqini qisqartirilgan rational songa tayanadi.

$a^{2/4}=a^{1/2}$

Negative base uchun unreduced denominator parity’ni mexanik o‘qish xavfli.

Even-denominator gate

Ratsional ko‘rsatkichning qisqartirilgan denominatori juft bo‘lsa negative base real emas.

$a^{m/n},\ n\text{ even}\Rightarrow a\ge0$

Negative exponent bo‘lsa a>0 talab qilinishi mumkin.

Odd-denominator permission

Qisqartirilgan denominator toq bo‘lsa negative base ham real qiymat berishi mumkin.

$(-8)^{1/3}=-2$

Odd root signni saqlaydi.

Principal-root absolute value

Juft ildiz powerni “bekor qilganda” sign yo‘qoladi va absolute value paydo bo‘ladi.

$\sqrt{x^2}=|x|$

Umumiy: n juft → sqrt[n](x^n)=|x|.

Odd-root cancellation

Toq ildiz va shu indeksdagi power real sohada to‘liq bekor bo‘ladi.

$\sqrt[3]{x^3}=x$

Barcha real x.

Product exponent law

Bir xil base ko‘paytirilsa ko‘rsatkichlar qo‘shiladi.

$a^r a^s=a^{r+s}$

Ifodalar real/aniqlangan bo‘lishi kerak.

Quotient exponent law

Bir xil base bo‘linsa ko‘rsatkichlar ayiriladi.

$a^r/a^s=a^{r-s}$

Denominator nonzero.

Power-of-power law

Darajaning darajasi ko‘rsatkichlar ko‘paytmasiga o‘tadi.

$(a^r)^s=a^{rs}$

Ratsional exponentsda negative base/domain uchun ehtiyotkorlik.

Negative exponent reciprocal

Negative exponent reciprocalga o‘tadi; ishorani almashtirmaydi.

$a^{-r}=1/a^r$

a≠0.

Radical product property

Bir xil indeksdagi real radicals mos shartlarda bitta radicalga birlashadi.

$\sqrt[n]{a}\sqrt[n]{b}=\sqrt[n]{ab}$

Even n da a,b≥0 kabi real-domain shartlari.

Radical quotient property

Bir xil indeksdagi quotient bitta radical quotientga birlashadi.

$\frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{a/b}$

Denominator root nonzero va barcha roots real.

Perfect-power extraction

Radikandni perfect n-th power × remainder ko‘rinishida yozing.

$\sqrt[n]{b^n c}$

n odd → b√[n]c; n even → |b|√[n]c.

Radical comparison by powers

Positive radicalsni umumiy musbat integer darajaga ko‘tarib radikand/powerlarni taqqoslash mumkin.

$u,v>0:\ u

N positive integer.

One-term rationalization

Denominator radicand exponentlarini indeksning to‘liq ko‘paytmasigacha to‘ldiring.

$1/\sqrt[n]{a^k}$

Missing power n-k multiplier tanlanadi.

Conjugate rationalization

Ikki hadli square-root denominator conjugate bilan difference of squaresga aylanadi.

$1/(\sqrt a+\sqrt b)$

Multiply by (√a-√b)/(√a-√b).

Cube-factor rationalization

Cube-root binomial denominator uchun u³±v³ factorizationdan foydalanish mumkin.

$(u\pm v)(u^2\mp uv+v^2)=u^3\pm v^3$

u,v cube-root expressions.

Like-radical collection

Radikallarni avval soddalashtirib, bir xil radical partli hadlarni coefficientlar bo‘yicha yig‘ing.

$c\sqrt[n]{r}+d\sqrt[n]{r}=(c+d)\sqrt[n]{r}$

Same simplified radical part.

Nested exponent multiplication

Ichma-ich roots rational exponents productiga aylanadi.

$\sqrt[m]{\sqrt[n]{x}}=x^{1/(mn)}$

Real-domain conditions.

Conjugate product under radicals

Nonnegative factors uchun sqrt(A)sqrt(B)=sqrt(AB); conjugate radicands AB ko‘pincha rational/perfect square bo‘ladi.

$\sqrt{p+q}\sqrt{p-q}=\sqrt{p^2-q^2}$

Har ikkala radicand ≥0.

Fixed-point infinite radical

Takroriy radical limit x bo‘lsa tail ham x ga teng deb fixed-point equation yoziladi.

$x=\sqrt{c x}$

x≥0 va iteration convergence/admissibility tekshiriladi.

Source integrity before solving

Prompt, options va marked answer bir-biriga mos kelmasa avval expressionni mustaqil soddalashtiring; extractionni taxmin bilan “tuzatmang”.

Invalid even roots va missing exponents canonical examplega kiritilmaydi.

Original-domain preservation

Algebraik cancellation yoki rationalization original expression domainini kengaytirmaydi.

Forbidden denominator/radicand points yakunda ham forbidden.

Formula kutubxonasi

Birlik kasr ko‘rsatkich

$$a^{1/n}=\sqrt[n]{a}$$
  • $n$ — ildiz indeksi, n≥2

Denominator ildiz indeksiga aylanadi.

Shart: sqrt[n](a) real bo‘lsa

Xususiy holatlar: n juft bo‘lsa a≥0.

Umumiy ratsional ko‘rsatkich

$$a^{m/n}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m$$
  • $m$ — numerator power
  • $n$ — denominator/index

Numerator daraja, denominator ildiz.

Shart: m/n qisqartirilgan va real expression aniqlangan

Xususiy holatlar: a>0 da barcha rational m/n uchun xavfsiz.

Manfiy ratsional daraja

$$a^{-m/n}=\frac{1}{a^{m/n}}$$
  • $a$ — base

Negative exponent reciprocal beradi.

Shart: a≠0 va a^{m/n} real

Xususiy holatlar: 16^{-1/2}=1/4.

Darajalar ko‘paytmasi

$$a^r a^s=a^{r+s}$$
  • $r,s$ — rational exponents

Bir xil base exponentlari qo‘shiladi.

Shart: Ifodalar real/aniqlangan; a>0 da universal

Xususiy holatlar: x^{1/2}x^{1/3}=x^{5/6}, x≥0.

Darajalar bo‘linmasi

$$\frac{a^r}{a^s}=a^{r-s}$$

Bir xil base exponentlari ayiriladi.

Shart: a≠0 va powers real

Xususiy holatlar: Denominator nonzero.

Darajaning darajasi

$$(a^r)^s=a^{rs}$$

Exponentlar ko‘payadi.

Shart: a>0 da barcha rational r,s uchun xavfsiz; boshqa holatda domain tekshiriladi

Xususiy holatlar: Negative base + fractional exponentsda mexanik qo‘llamang.

Ko‘paytmaning darajasi

$$(ab)^r=a^r b^r$$

Power product bo‘yicha tarqaladi.

Shart: Real expressions aniqlangan; a,b>0 da xavfsiz

Xususiy holatlar: Even-denominator rootsda negative factorsni alohida birlashtirish xavfli.

Kasrning darajasi

$$(a/b)^r=\frac{a^r}{b^r}$$

Power numerator va denominatorga tarqaladi.

Shart: b≠0 va real powers aniqlangan

Xususiy holatlar: Domain saqlanadi.

Nol daraja

$$a^0=1$$

Nonzero base zero-power 1.

Shart: a≠0

Xususiy holatlar: 0^0 bu kursda aniqlanmagan.

Juft principal cancellation

$$\sqrt[n]{a^n}=|a|$$
  • $n$ — juft musbat integer

Juft principal root signni nonnegative qiladi.

Shart: n juft

Xususiy holatlar: sqrt(a²)=|a|.

Toq root cancellation

$$\sqrt[n]{a^n}=a$$
  • $n$ — toq musbat integer

Toq root signni saqlaydi.

Shart: n toq

Xususiy holatlar: cubert(a³)=a.

Power after root

$$(\sqrt[n]{a})^n=a$$

Principal rootni ayni n-darajaga ko‘tarish radikandni qaytaradi.

Shart: sqrt[n](a) real

Xususiy holatlar: Even n da a≥0.

Radikal product property

$$\sqrt[n]{a}\,\sqrt[n]{b}=\sqrt[n]{ab}$$

Bir xil indeksli radicals ko‘payadi.

Shart: Ikkala root va product root real; even n da odatda a,b≥0

Xususiy holatlar: Odd n negative factorsga ham ruxsat beradi.

Radikal quotient property

$$\frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac ab}$$

Bir xil indeksli radical quotient.

Shart: sqrt[n](b)≠0 va roots real

Xususiy holatlar: b≠0 va domain.

Perfect-power extraction — juft

$$\sqrt[n]{b^n c}=|b|\sqrt[n]{c}$$

Perfect n-th power tashqariga absolute value bilan chiqadi.

Shart: n juft, expression real

Xususiy holatlar: b signi noma’lum bo‘lsa |b| shart.

Perfect-power extraction — toq

$$\sqrt[n]{b^n c}=b\sqrt[n]{c}$$

Odd rootda perfect power sign bilan chiqadi.

Shart: n toq

Xususiy holatlar: Negative b mumkin.

O‘xshash radikallar

$$c\sqrt[n]{r}+d\sqrt[n]{r}=(c+d)\sqrt[n]{r}$$

Coefficientlar yig‘iladi.

Shart: Radical qismlar aynan bir xil

Xususiy holatlar: Oldin radicalsni soddalashtiring.

Kvadrat ildizli bir hadli rationalization

$$\frac{1}{\sqrt a}=\frac{\sqrt a}{a}$$

Denominatordagi sqrt yo‘qoladi.

Shart: a>0

Xususiy holatlar: Coefficient bo‘lsa shu usul scale qilinadi.

Umumiy n-root rationalization

$$\frac{1}{\sqrt[n]{a^k}}\cdot\frac{\sqrt[n]{a^{n-k}}}{\sqrt[n]{a^{n-k}}}=\frac{\sqrt[n]{a^{n-k}}}{a}$$

Exponentni n gacha to‘ldirib maxrajni rational qiladi.

Shart: 0<k<n, denominator real va nonzero

Xususiy holatlar: Sign/domain n ga bog‘liq.

Conjugate identity

$$(u+v)(u-v)=u^2-v^2$$

Ikki hadli square-root denominatorni rational qiladi.

Shart: u,v real expressions

Xususiy holatlar: u=√a,v=√b → a-b.

Kub yig‘indisi

$$(u+v)(u^2-uv+v^2)=u^3+v^3$$

Cube-root binomial rationalization multiplier.

Shart: Algebraic identity

Xususiy holatlar: Difference versiyasi alohida.

Kub ayirmasi

$$(u-v)(u^2+uv+v^2)=u^3-v^3$$

Cube-root difference denominator rationalization.

Shart: Algebraic identity

Xususiy holatlar: u³-v³ denominator rational bo‘lishi mumkin.

Ichma-ich ildiz

$$\sqrt[m]{\sqrt[n]{a}}=a^{1/(mn)}$$

Indekslar rational exponentda ko‘payadi.

Shart: Barcha nested roots real

Xususiy holatlar: Even rootsda a≥0 talab qilinadi.

Musbat radical comparison

$$u
  • $N$ — musbat integer

Nonnegative sonlarda positive integer power tartibni saqlaydi.

Shart: u,v≥0

Xususiy holatlar: N ni indekslar LCMiga mos tanlang.

Masofa ko‘rinishidagi square root

$$\sqrt{a^2}=|a|$$

Square root square’ni absolute value qiladi.

Shart: a real

Xususiy holatlar: a≥0 bo‘lsa |a|=a; a<0 bo‘lsa |a|=-a.

Cheksiz sqrt fixed point

$$x=\sqrt{c x}\Longrightarrow x^2=cx$$
  • $c$ — musbat constant

Tail self-similarity fixed-point equation beradi.

Shart: Nested radical convergent va x≥0

Xususiy holatlar: Algebraic rootsdan faqat admissible/convergent root tanlanadi.

Teoremalar va isbotlar

📐 Ratsional ko‘rsatkich–ildiz teoremasi

m/n qisqartirilgan rational son bo‘lsin. Real expression mavjud bo‘lsa a^(m/n)=sqrt[n](a^m)=(sqrt[n](a))^m.

Denominator “qaysi ildiz”, numerator “qaysi daraja”ni aytadi.

Isbotni ko'rsatish

Berilgan: m/n qisqartirilgan va expression real

Isbotlash kerak: a^(m/n)=sqrt[n](a^m)=(sqrt[n](a))^m

  1. a^(1/n)=sqrt[n](a) deb olinadi.
  2. Har ikki tomonni m-darajaga ko‘tarsak (a^(1/n))^m=(sqrt[n](a))^m.
  3. Exponent law bilan chap tomon a^(m/n).
  4. Root-power notation bo‘yicha (sqrt[n](a))^m=sqrt[n](a^m), real-domain shartlari saqlangan holatda.

Rational exponent va radical notation ekvivalent. ∎

📐 Principal even-root absolute-value teoremasi

Juft n uchun sqrt[n](a^n)=|a|.

Principal even root nonnegative bo‘lishi shart; a manfiy bo‘lsa -a olinadi.

Isbotni ko'rsatish

Berilgan: n juft, a real

Isbotlash kerak: sqrt[n](a^n)=|a|

  1. a^n=(-a)^n juft n uchun.
  2. |a|≥0 va |a|^n=a^n.
  3. Principal n-th root a^n ning nonnegative n-th rootini tanlaydi.
  4. Shuning uchun sqrt[n](a^n)=|a|.

Even principal root perfect powerdan absolute value chiqaradi. ∎

📐 Positive-base exponent laws teoremasi

a>0 va rational r,s uchun product, quotient, power-of-power, product-to-power va quotient-to-power qonunlari odatdagi ko‘rinishda ishlaydi.

Positive base barcha required rootsni real qiladi va branch ambiguityni olib tashlaydi.

Isbotni ko'rsatish

Berilgan: a>0; r,s rational

Isbotlash kerak: Asosiy exponent laws ratsional exponentsda ishlashi

  1. r=p/n va s=q/n kabi umumiy denominator tanlash mumkin.
  2. a^r=(a^(1/n))^p va a^s=(a^(1/n))^q.
  3. Integer exponent product law bilan ko‘paytma (a^(1/n))^(p+q)=a^(r+s).
  4. Quotient va power-of-power ayni usul bilan keladi.
  5. a>0 barcha involved rootsni real va nonzero qiladi.

Positive-base rational exponent algebra consistent. ∎

📐 Radikal product/quotient teoremasi

Bir xil indeksli real radicals uchun mos domain shartlarida product va quotient radical ichida birlashtirilishi mumkin.

nth-root operation compatible productsga taqsimlanadi.

Isbotni ko'rsatish

Berilgan: nth roots real

Isbotlash kerak: Product/quotient root properties

  1. u=sqrt[n](a), v=sqrt[n](b) deb oling.
  2. (uv)^n=u^n v^n=ab.
  3. Mos domain va principal-root sign shartlarida uv sqrt[n](ab) ning kerakli real ildizidir.
  4. Quotient uchun (u/v)^n=a/b va v≠0.

Radikal product va quotient xossalari domain bilan isbotlandi. ∎

📐 Conjugate rationalization teoremasi

Square-root binomial denominatorni uning conjugate’iga ko‘paytirish difference of squares orqali radical-free denominator beradi.

Cross terms bir-birini bekor qiladi.

Isbotni ko'rsatish

Berilgan: D=sqrt(a)+sqrt(b) yoki sqrt(a)-sqrt(b)

Isbotlash kerak: Conjugate maxrajni radical-free qiladi

  1. Conjugate D*=sqrt(a)-sqrt(b) yoki sqrt(a)+sqrt(b).
  2. D·D*=(sqrt(a))²-(sqrt(b))²=a-b.
  3. Kasr qiymatini saqlash uchun numerator va denominatorni bir xil nonzero D* ga ko‘paytiramiz.
  4. Natijada denominator rational algebraic expression a-b bo‘ladi.

Conjugate rationalization difference-of-squaresga tayanadi. ∎

Yechilgan misollar

oson $8^{2/3}$ ni hisoblang.

💡 Maslahat: Avval cube root, keyin square.

  1. 8^{2/3}=(\sqrt[3]{8})^2.
  2. \sqrt[3]{8}=2.
  3. 2^2=4.

✅ Javob: $4$

Nega bu usul ishlaydi: Rational exponentning denominator qismi root, numerator qismi power.

Muqobil usul: $8^{2/3}=\sqrt[3]{8^2}=\sqrt[3]{64}=4$.

⚠️ Root first sonlarni kichik saqlaydi.

oson $16^{-3/2}$ ni hisoblang.

💡 Maslahat: Negative exponentni reciprocalga aylantiring.

  1. 16^{-3/2}=1/16^{3/2}.
  2. 16^{3/2}=(\sqrt{16})^3=4^3=64.
  3. Natija 1/64.

✅ Javob: $\frac1{64}$

Nega bu usul ishlaydi: Negative exponent reciprocal, fractional exponent esa root-power beradi.

⚠️ Minus exponent qiymatni manfiy qilmaydi.

ortacha $\frac{(a^2b^3)^{-2}(ab^{-1})^3}{a^{-2}b^3}$ ni soddalashtiring.

💡 Maslahat: Har base exponentini alohida yig‘ing/ayiring.

  1. (a²b³)^(-2)=a^(-4)b^(-6).
  2. (ab^(-1))³=a³b^(-3).
  3. Numerator a^(-1)b^(-9).
  4. Denominatorga bo‘lish: a^(-1-(-2))b^(-9-3)=ab^(-12).

✅ Javob: $ab^{-12}$

Nega bu usul ishlaydi: Bir xil base uchun exponent arithmetic ishlaydi.

⚠️ Source 7844 bilan mos.

ortacha $\left(\frac{a^2b^{-1}}{a^3b^{-1}}\right)^{-2}$ ni soddalashtiring.

💡 Maslahat: Ichki quotientda b lar bekor bo‘ladi.

  1. Ichkarida a^(2-3)b^(-1-(-1))=a^(-1).
  2. (a^(-1))^(-2)=a².

✅ Javob: $a^2$

Nega bu usul ishlaydi: Quotient va power-of-power ketma-ket ishlaydi.

⚠️ Source 7843 marked `a^{-2}b^{-4}` noto‘g‘ri; b butunlay qisqaradi.

oson $a^4\cdot a^3:a^5$ ni soddalashtiring.

💡 Maslahat: Exponentlarni 4+3-5 qiling.

  1. a⁴·a³/a⁵=a^(4+3-5).
  2. 4+3-5=2.

✅ Javob: $a^2$

Nega bu usul ishlaydi: Bir xil base product/quotient law.

⚠️ Source 7848 `1` deb belgilagan; canonical javob a².

ortacha $\frac{a^{-1}-a}{a^{-1}}$ ni soddalashtiring.

💡 Maslahat: Butun numeratorni a^{-1} ga bo‘ling.

  1. (a^(-1)-a)/a^(-1)=1-a/a^(-1).
  2. a/a^(-1)=a².
  3. Natija 1-a².

✅ Javob: $1-a^2$

Nega bu usul ishlaydi: Kasrni term-by-term bo‘lish yoki denominator reciprocaliga ko‘paytirish mumkin.

Muqobil usul: $a(a^{-1}-a)=1-a²$.

⚠️ Source 7849 `a²` deb belgilagan.

murakkab $\left[b(ab^{-1})^{-1}a^{-1}\right]^5$ ni soddalashtiring.

💡 Maslahat: Avval bracket ichini soddalashtiring.

  1. (ab^(-1))^(-1)=a^(-1)b.
  2. b·a^(-1)b·a^(-1)=a^(-2)b².
  3. Beshinchi daraja: a^(-10)b^10.

✅ Javob: $a^{-10}b^{10}$

Nega bu usul ishlaydi: Power-of-product va power-of-power.

Muqobil usul: $\frac{b^{10}}{a^{10}}$.

⚠️ Source 7850 variantlarining hech biri mos emas.

ortacha $(x^{-1}y^{-1})^4(xy)$ ni soddalashtiring.

💡 Maslahat: Bracket exponentini tarqating.

  1. (x^(-1)y^(-1))⁴=x^(-4)y^(-4).
  2. xy ga ko‘paytirish exponentlarni +1 qiladi.
  3. x^(-3)y^(-3).

✅ Javob: $x^{-3}y^{-3}$

Nega bu usul ishlaydi: Product exponent laws.

Muqobil usul: $1/(x³y³)$.

⚠️ Source 7851 marked `x` noto‘g‘ri.

oson $(xy^{-1})^{-2}(xy^{-1})^3$ ni soddalashtiring.

💡 Maslahat: Bir xil base-expression exponentlari qo‘shiladi.

  1. (xy^(-1))^(-2+3)=xy^(-1).

✅ Javob: $xy^{-1}=\frac{x}{y}$

Nega bu usul ishlaydi: Bir xil butun expression base sifatida ko‘riladi.

⚠️ Source 7852 A=`xy` emas; D mazmunan to‘g‘ri.

oson $\sqrt{x}\sqrt{x}\sqrt{x}$ ni daraja ko‘rinishida yozing, $x\ge0$.

💡 Maslahat: Har sqrt x = x^(1/2).

  1. x^(1/2)x^(1/2)x^(1/2)=x^(3/2).

✅ Javob: $x^{3/2}$

Nega bu usul ishlaydi: Bir xil base fractional exponents qo‘shiladi.

Muqobil usul: $x\sqrt{x}$.

⚠️ Source 7854 `x^{1/2}` deb belgilagan; C to‘g‘ri.

ortacha $x\sqrt{x}\sqrt[3]{x}$ ni daraja ko‘rinishida yozing, $x>0$.

💡 Maslahat: 1+1/2+1/3 ni qo‘shing.

  1. x=x¹, sqrt(x)=x^(1/2), cubert(x)=x^(1/3).
  2. 1+1/2+1/3=6/6+3/6+2/6=11/6.

✅ Javob: $x^{11/6}$

Nega bu usul ishlaydi: Radicals rational exponentsga o‘tkaziladi.

⚠️ Source 7855 variantlarida canonical exponent yo‘q.

ortacha $x^2\sqrt{x}\sqrt[3]{x}\sqrt[4]{x}$ ni bitta darajaga keltiring, $x>0$.

💡 Maslahat: 2+1/2+1/3+1/4.

  1. LCM 12.
  2. 2=24/12, 1/2=6/12, 1/3=4/12, 1/4=3/12.
  3. Jami 37/12.

✅ Javob: $x^{37/12}$

Nega bu usul ishlaydi: Common denominator exponent additionni aniq qiladi.

⚠️ Source 7856 variantlarida to‘g‘ri javob yo‘q.

oson $\sqrt{\sqrt{x}}$ ni rational exponentda yozing, $x\ge0$.

💡 Maslahat: Har square root exponentni 1/2 ga ko‘paytiradi.

  1. sqrt(x)=x^(1/2).
  2. sqrt(sqrt(x))=(x^(1/2))^(1/2)=x^(1/4).

✅ Javob: $x^{1/4}$

Nega bu usul ishlaydi: Nested root = exponent multiplication.

Muqobil usul: $\sqrt[4]{x}$.

oson $\sqrt{x^2}$ ni soddalashtiring, $x$ real.

💡 Maslahat: Principal square root nonnegative.

  1. x² ning ikki square rooti ±x bo‘lishi mumkin, lekin radical belgisi principal nonnegative rootni tanlaydi.
  2. Shuning uchun sqrt(x²)=|x|.

✅ Javob: $|x|$

Nega bu usul ishlaydi: Natija har doim nonnegative bo‘lishi kerak.

⚠️ `sqrt(x²)=x` faqat x≥0 da.

oson $\sqrt[3]{-27}$ ni hisoblang.

💡 Maslahat: Toq ildiz manfiy radikandga ruxsat beradi.

  1. (-3)³=-27.
  2. Shuning uchun cubert(-27)=-3.

✅ Javob: $-3$

Nega bu usul ishlaydi: Odd root signni saqlaydi.

⚠️ Even root bilan aralashtirmang.

ortacha Agar $a<0<b$ bo‘lsa, $\sqrt[3]{a^3}+\sqrt{b^2}-\sqrt{a^2}-\sqrt{(a-b)^2}$ ni soddalashtiring.

💡 Maslahat: Odd root va principal square rootni alohida ishlating.

  1. cubert(a³)=a.
  2. b>0 → sqrt(b²)=b.
  3. a<0 → sqrt(a²)=|a|=-a.
  4. a-b<0 → sqrt((a-b)²)=|a-b|=b-a.
  5. a+b-(-a)-(b-a)=3a.

✅ Javob: $3a$

Nega bu usul ishlaydi: Juft roots absolute value, toq root esa signni saqlaydi.

⚠️ Source 7866 A bilan mos va sign-domain uchun yaxshi canonical misol.

oson $27^{2/3}-(1/16)^{-1/2}$ ni hisoblang.

💡 Maslahat: Har hadni alohida root-power ko‘rinishida hisoblang.

  1. 27^(2/3)=(cubert27)²=3²=9.
  2. (1/16)^(-1/2)=1/sqrt(1/16)=1/(1/4)=4.
  3. 9-4=5.

✅ Javob: $5$

Nega bu usul ishlaydi: Fractional va negative exponent qoidalari birga ishlaydi.

⚠️ Source 7880 bilan mos.

ortacha $\frac{\sqrt{40}+\sqrt{90}}{\sqrt{160}+\sqrt{360}}$ ni soddalashtiring.

💡 Maslahat: Har radikalni √10 ga keltiring.

  1. sqrt40=2sqrt10, sqrt90=3sqrt10.
  2. sqrt160=4sqrt10, sqrt360=6sqrt10.
  3. Kasr 5sqrt10/(10sqrt10)=1/2.

✅ Javob: $\frac12$

Nega bu usul ishlaydi: Like radicals common factorga keltiriladi.

⚠️ Source 7885 bilan mos.

ortacha $\frac{\sqrt[3]{48}+\sqrt[3]{162}}{\sqrt[3]{750}}$ ni soddalashtiring.

💡 Maslahat: Har radikanddan perfect cube factor ajrating.

  1. cubert48=cubert(8·6)=2cubert6.
  2. cubert162=cubert(27·6)=3cubert6.
  3. cubert750=cubert(125·6)=5cubert6.
  4. Natija (2+3)/5=1.

✅ Javob: $1$

Nega bu usul ishlaydi: Perfect cube extraction barcha hadlarda bir xil radical part yaratadi.

⚠️ Source 7886 bilan mos.

ortacha $\sqrt{6+\sqrt{11}}\,\sqrt{6-\sqrt{11}}$ ni hisoblang.

💡 Maslahat: Ikkala radikand nonnegative; productni bitta sqrt ichiga birlashtiring.

  1. (6+sqrt11)(6-sqrt11)=36-11=25.
  2. sqrt25=5.

✅ Javob: $5$

Nega bu usul ishlaydi: Conjugate product radical ichida difference of squares beradi.

⚠️ Source 7896 marked `√5` noto‘g‘ri; B=5 canonical.

ortacha $\sqrt{\sqrt{43}+\sqrt{11}}\,\sqrt{\sqrt{43}-\sqrt{11}}$ ni hisoblang.

💡 Maslahat: Product radicandlar conjugate.

  1. Product = sqrt((sqrt43)²-(sqrt11)²).
  2. =sqrt(43-11)=sqrt32=4sqrt2.

✅ Javob: $4\sqrt2$

Nega bu usul ishlaydi: Radical product + difference of squares.

⚠️ Source 7897 marked 2 noto‘g‘ri; option `√32` mazmunan to‘g‘ri.

ortacha $\sqrt{4\sqrt2-\sqrt5}\,\sqrt{4\sqrt2+\sqrt5}$ ni hisoblang.

💡 Maslahat: Conjugate radicands.

  1. Product = sqrt((4sqrt2)²-(sqrt5)²).
  2. (4sqrt2)²=32, (sqrt5)²=5.
  3. sqrt(27)=3sqrt3.

✅ Javob: $3\sqrt3$

Nega bu usul ishlaydi: Conjugate structure perfect simplification beradi.

⚠️ Source 7898 marked 3 noto‘g‘ri; `√27` option mazmunan to‘g‘ri.

oson $\frac{\sqrt5-2}{\sqrt{9-4\sqrt5}}$ ni soddalashtiring.

💡 Maslahat: 9-4√5 ni kvadrat sifatida taning.

  1. 9-4sqrt5=(sqrt5-2)².
  2. sqrt((sqrt5-2)²)=|sqrt5-2|.
  3. sqrt5>2, shuning uchun |sqrt5-2|=sqrt5-2.
  4. Kasr 1.

✅ Javob: $1$

Nega bu usul ishlaydi: Principal root absolute value bilan ishlaydi.

⚠️ Source 7904 bilan mos.

ortacha $\sqrt[3]{26}$ va $\sqrt5$ ni taqqoslang.

💡 Maslahat: Ikkalasi musbat; 6-darajaga ko‘taring.

  1. u=cubert26, v=sqrt5.
  2. u^6=26²=676.
  3. v^6=5³=125.
  4. 676>125 → u>v.

✅ Javob: $\sqrt[3]{26}>\sqrt5$

Nega bu usul ishlaydi: Musbat sonlarda bir xil positive power tartibni saqlaydi.

⚠️ Source 7905 prompt/options extractioni chalkash, lekin shu comparison canonical.

ortacha $\sqrt[4]{4}$ va $\sqrt[3]{3}$ ni taqqoslang.

💡 Maslahat: 12-darajaga ko‘taring.

  1. (fourthroot4)^12=4³=64.
  2. (cubert3)^12=3⁴=81.
  3. 64<81.

✅ Javob: $\sqrt[4]{4}<\sqrt[3]{3}$

Nega bu usul ishlaydi: LCM(4,3)=12 common comparison power beradi.

⚠️ Source 7906 A mazmunan to‘g‘ri.

oson $\frac{2}{\sqrt2}$ ni ratsionallashtiring.

💡 Maslahat: √2/√2 ga ko‘paytiring.

  1. 2/sqrt2 · sqrt2/sqrt2 = 2sqrt2/2.
  2. =sqrt2.

✅ Javob: $\sqrt2$

Nega bu usul ishlaydi: Denominator square root o‘zi bilan ko‘payganda rational 2 bo‘ladi.

⚠️ Source 7914 marked `√8` noto‘g‘ri; C expression √2 ga teng.

oson $\frac1{\sqrt5}$ ni ratsionallashtiring.

💡 Maslahat: √5/√5 ga ko‘paytiring.

  1. 1/sqrt5 · sqrt5/sqrt5 = sqrt5/5.

✅ Javob: $\frac{\sqrt5}{5}$

Nega bu usul ishlaydi: Equivalent fraction denominatorni rational qiladi.

⚠️ Source 7915 B to‘g‘ri; A marked noto‘g‘ri.

oson $\frac5{\sqrt4}$ ni soddalashtiring.

💡 Maslahat: √4 ni avval hisoblang.

  1. sqrt4=2.
  2. 5/sqrt4=5/2.

✅ Javob: $\frac52$

Nega bu usul ishlaydi: Perfect square denominatorni rationalization qilmasdan to‘g‘ridan hisoblash mumkin.

⚠️ Source 7916 D to‘g‘ri; A marked noto‘g‘ri.

ortacha $\frac1{\sqrt3-\sqrt2}$ ni ratsionallashtiring.

💡 Maslahat: Conjugate √3+√2.

  1. Kasrni (sqrt3+sqrt2)/(sqrt3+sqrt2) ga ko‘paytiring.
  2. Denominator 3-2=1.
  3. Numerator sqrt3+sqrt2.

✅ Javob: $\sqrt3+\sqrt2$

Nega bu usul ishlaydi: Conjugate denominatorni difference of squaresga aylantiradi.

⚠️ Source 7918 marked A noto‘g‘ri; B canonical.

ortacha $\frac{2}{\sqrt2-\sqrt3}$ ni ratsionallashtiring.

💡 Maslahat: Conjugate √2+√3; denominator manfiy chiqadi.

  1. 2/(sqrt2-sqrt3) · (sqrt2+sqrt3)/(sqrt2+sqrt3).
  2. Denominator 2-3=-1.
  3. Natija -2(sqrt2+sqrt3).

✅ Javob: $-2(\sqrt2+\sqrt3)$

Nega bu usul ishlaydi: Conjugate signni ham to‘g‘ri saqlash kerak.

⚠️ Source 7922 variantlarida canonical form yo‘q.

ortacha $\frac{a+8}{\sqrt[3]{a^2}-2\sqrt[3]{a}+4}$ ni soddalashtiring.

💡 Maslahat: t=∛a deb oling va a+8=t³+2³ ni factorlang.

  1. t=cubert(a).
  2. a+8=t³+8=(t+2)(t²-2t+4).
  3. Denominator aynan t²-2t+4.
  4. Qisqartirib t+2=cubert(a)+2.

✅ Javob: $\sqrt[3]{a}+2$

Nega bu usul ishlaydi: Sum of cubes denominator strukturasini bekor qiladi.

⚠️ Source 7926 B to‘g‘ri; A marked `sixthroot(a)+2` noto‘g‘ri.

ortacha $\frac{a^{-1}-b^{-1}}{a^{-1}+b^{-1}}$ ni soddalashtiring, $ab\ne0$.

💡 Maslahat: Numerator va denominatorni ab ga ko‘paytiring.

  1. 1/a-1/b=(b-a)/(ab).
  2. 1/a+1/b=(a+b)/(ab).
  3. Quotient (b-a)/(a+b), a+b≠0.

✅ Javob: $\frac{b-a}{a+b}$

Nega bu usul ishlaydi: Negative exponents reciprocal sifatida ochiladi.

⚠️ Source 7934 variantlari canonical expressionga mos emas.

oson $\sqrt[4]{97\cdot103+9}$ ni hisoblang.

💡 Maslahat: 97·103=(100-3)(100+3).

  1. 97·103=10000-9=9991.
  2. +9 → 10000.
  3. 10000=10⁴.
  4. fourthroot(10⁴)=10.

✅ Javob: $10$

Nega bu usul ishlaydi: Difference-of-squares yashirin perfect fourth power beradi.

⚠️ Source 7943 bilan mos.

ortacha $\sqrt{220\cdot230+25}$ ni hisoblang.

💡 Maslahat: 220·230=(225-5)(225+5).

  1. 220·230=225²-25.
  2. +25 →225².
  3. Principal sqrt(225²)=225.

✅ Javob: $225$

Nega bu usul ishlaydi: Conjugate-number product perfect square yaratadi.

⚠️ Source 7944 marked 15 noto‘g‘ri.

ortacha $\sqrt[4]{99\cdot10101+1}$ ni hisoblang.

💡 Maslahat: 99·10101 ni hisoblab perfect powerni taning.

  1. 10101·99=10101(100-1)=1010100-10101=999999.
  2. +1=1000000=10⁶.
  3. fourthroot(10⁶)=10^(6/4)=10^(3/2)=10sqrt10.

✅ Javob: $10\sqrt{10}$

Nega bu usul ishlaydi: Integer radicandni power formaga keltirish rational exponent hisobidir.

⚠️ Source 7945 marked 100 noto‘g‘ri.

murakkab $\sqrt{64\sqrt{64\sqrt{64\cdots}}}=x$ convergent va musbat bo‘lsa, $x^3$ ni toping.

💡 Maslahat: Tail yana x ga teng.

  1. x=sqrt(64x).
  2. x≥0.
  3. Square: x²=64x → x(x-64)=0.
  4. Positive nontrivial fixed point x=64.
  5. x³=64³=262144.

✅ Javob: $262144$

Nega bu usul ishlaydi: Infinite radical self-similarity fixed-point equation beradi.

Muqobil usul: Iteration convergence alohida tekshirilishi mumkin.

⚠️ Source 7949 x³ uchun 64 deb belgilagan; bu x ning o‘zi.

oson $x\cdot x^2\cdot x\sqrt{x}$ ni bitta darajaga keltiring, $x\ge0$.

💡 Maslahat: Exponentlar 1+2+1+1/2.

  1. x·x²·x·sqrt(x)=x^(1+2+1+1/2).
  2. Jami 9/2.

✅ Javob: $x^{9/2}$

Nega bu usul ishlaydi: Bir xil base exponentlari qo‘shiladi.

Muqobil usul: $x^{4.5}$.

⚠️ Source 7961 A va B bir xil qiymatni ifodalaydi; test variantlari duplicate.

oson $x\sqrt{x}\sqrt{\sqrt{x}}$ ni bitta darajaga keltiring, $x\ge0$.

💡 Maslahat: sqrt(sqrt(x))=x^(1/4).

  1. x=x¹.
  2. sqrt(x)=x^(1/2).
  3. sqrt(sqrt(x))=x^(1/4).
  4. 1+1/2+1/4=7/4.

✅ Javob: $x^{7/4}$

Nega bu usul ishlaydi: Nested radical rational exponentga aylanadi.

⚠️ Source 7962 variantlarida 7/4 yo‘q.

oson $8^{1/3}+16^{1/4}$ ni hisoblang.

💡 Maslahat: Perfect cube va fourth power.

  1. cubert8=2.
  2. fourthroot16=2.
  3. 2+2=4.

✅ Javob: $4$

Nega bu usul ishlaydi: Fractional powers rootlar sifatida hisoblanadi.

⚠️ Source 7963 bilan mos.

oson $81^{1/4}+27^{1/4}$ ni soddalashtiring.

💡 Maslahat: Har hadni 3 asosida yozing.

  1. 81=3⁴ →81^(1/4)=3.
  2. 27=3³ →27^(1/4)=3^(3/4).
  3. Yig‘indi 3+3^(3/4).

✅ Javob: $3+3^{3/4}$

Nega bu usul ishlaydi: Har radicand perfect fourth power bo‘lishi shart emas.

⚠️ Source 7964 marked 30 va barcha integer variantlar noto‘g‘ri.

oson $125^{1/3}-343^{1/3}$ ni hisoblang.

💡 Maslahat: 125=5³, 343=7³.

  1. cubert125=5.
  2. cubert343=7.
  3. 5-7=-2.

✅ Javob: $-2$

Nega bu usul ishlaydi: Odd roots exact integerlarga tushadi.

⚠️ Source 7965 variantlarida -2 yo‘q.

ortacha $\sqrt[4]{-16}$ ning real qiymatini toping.

💡 Maslahat: Index juft.

  1. Juft indeksli real root uchun radikand ≥0 bo‘lishi kerak.
  2. -16<0.
  3. Shuning uchun real qiymat mavjud emas.

✅ Javob: Real sonlarda aniqlanmagan

Nega bu usul ishlaydi: Even-index domain gate har qanday algebraik manipulationdan oldin keladi.

⚠️ Source bankdagi 7917 kabi negative square-root records canonical real darsda shu sabab yaroqsiz.

murakkab $\sqrt{(2-\sqrt5)^3}$ real sonmi?

💡 Maslahat: 2-√5 ishorasini tekshiring.

  1. sqrt5>2, demak 2-sqrt5<0.
  2. Negative sonning 3-darajasi ham negative.
  3. Square root negative radikand ustida real emas.

✅ Javob: Real sonlarda aniqlanmagan

Nega bu usul ishlaydi: Outer even root radikand signi hal qiluvchi.

Muqobil usul: Complex numbersda boshqa talqin mavjud, lekin bu mavzu real algebra doirasida.

⚠️ Source 7868 real numeric javob variantlari beradi; record real-domain jihatdan buzilgan.

Umumiy xatolar

❌ Kasr ko‘rsatkich denominatorini daraja, numeratorini ildiz deb almashtirish.

Rational exponentda denominator ildiz indeksidir.

✅ a^(m/n)=sqrt[n](a^m) ni ishlating.

8^(2/3)=(cubert8)².

❌ Negative exponent qiymatni manfiy qiladi deb o‘ylash.

Minus exponent reciprocalni bildiradi.

✅ a^(-r)=1/a^r.

16^(-1/2)=1/4.

❌ Bir xil base ko‘paytmada exponentlarni ko‘paytirish.

Product law exponentlarni qo‘shadi.

✅ a^r a^s=a^(r+s).

x^(1/2)x^(1/3)=x^(5/6).

❌ Quotientda exponentlarni qo‘shish.

Denominator exponenti ayriladi.

✅ a^r/a^s=a^(r-s).

a⁴a³/a⁵=a².

❌ (a^r)^s da exponentlarni qo‘shish.

Power-of-power exponentlarni ko‘paytiradi.

✅ (a^r)^s=a^(rs).

(a^(-1))^(-2)=a².

❌ sqrt(x²)=x deb barcha real x uchun yozish.

Principal square root nonnegative.

✅ sqrt(x²)=|x|.

x=-3 da sqrt9=3≠-3.

❌ Juft ildiz ostida negative sonni real deb hisoblash.

Even-index root negative radikandda real emas.

✅ Avval radikand≥0 domainini tekshiring.

fourthroot(-16) real emas.

❌ Toq ildizda manfiy radikandni taqiqlash.

Odd-index roots barcha real radikandlarda mavjud.

✅ cubert(-27)=-3.

Cube root signni saqlaydi.

❌ Perfect even powerni radikaldan signsiz chiqarish.

Even principal root absolute value beradi.

✅ sqrt[n](b^n)=|b|, n even.

sqrt(a²)=|a|.

❌ sqrt(a+b)=sqrt(a)+sqrt(b) deb ajratish.

Root product/quotientga tarqalishi mumkin, sumga umumiy tarqalmaydi.

✅ Sumni factor/perfect square structure bo‘lsa boshqa usul bilan soddalashtiring.

sqrt(9+16)=5, 3+4=7 emas.

❌ sqrt(a)sqrt(b)=sqrt(ab) ni domain tekshirmasdan ishlatish.

Even roots real bo‘lishi uchun individual radikandlar shartga mos bo‘lishi kerak.

✅ Real-domain shartlarini yozing.

sqrt(-1)sqrt(-1) real algebra doirasida mavjud emas.

❌ Radikallarni qo‘shishda radikandlarni qo‘shish.

Faqat like radicals coefficient bo‘yicha yig‘iladi.

✅ Oldin soddalashtiring, keyin bir xil radical partlarni birlashtiring.

2sqrt3+5sqrt3=7sqrt3.

❌ Rationalizationda faqat denominatorni ko‘paytirish.

Bu kasr qiymatini o‘zgartiradi.

✅ Numerator va denominatorni bir xil nonzero multiplierga ko‘paytiring.

1/sqrt5 · sqrt5/sqrt5.

❌ Ikki hadli sqrt denominatorni bitta sqrt bilan ko‘paytirish.

Cross radical terms qolib ketadi.

✅ Conjugate ishlating.

1/(sqrt3-sqrt2) → sqrt3+sqrt2.

❌ Soddalashtirgandan keyin original denominator restrictionni unutish.

Cancellation domainni kengaytirmaydi.

✅ Original domainni yakunda saqlang.

Keyingi irrational/rational expressionsda muhim.

❌ Turli indeksli radicalsni decimal taxmin bilan solishtirishga majbur bo‘lish.

Exact comparison common positive power bilan mumkin.

✅ Indekslar LCMiga mos darajaga ko‘taring.

fourthroot4 vs cubert3: 64<81.

❌ Infinite radical fixed-point equationdagi barcha rootsni qabul qilish.

Limit nonnegative va iterationga mos bo‘lishi kerak.

✅ Admissibility/convergence shartini tekshiring.

x=sqrt(64x) da positive fixed point x=64.

❌ Extraction-buzilgan savolni variantga qarab taxminan “tiklash”.

Promptning o‘zi boshqa matematika bo‘lishi mumkin.

✅ Expressionni aynan saqlangan matn bo‘yicha tekshiring; ishonchsiz recordni QA flag qiling.

7853,7874–7876,7933.

❌ Marked correct_optionni algebraik tekshiruvsiz qabul qilish.

Bankda ko‘plab marked answer xatolari bor.

✅ Independent simplification va substitution bilan tekshiring.

7843,7848,7854,7896,7914–7918,7944.

❌ Kasr ko‘rsatkichni qisqartirmasdan denominator paritydan domain chiqarish.

Rational numberning qiymati uning qisqartirilgan ko‘rinishiga bog‘liq.

✅ Avval m/n ni lowest termsga keltiring.

a^(2/4)=a^(1/2).

Noto'g'ri tasavvurlar

Ratsional ko‘rsatkich faqat positive base uchun mavjud.

Positive base eng xavfsiz umumiy holat; odd denominatorli qisqartirilgan exponentlarda negative base ham real bo‘lishi mumkin.

Har qanday negative base fractional exponenti real.

Even denominatorli qisqartirilgan exponentlarda negative base real emas.

sqrt(a²)=a.

To‘g‘ri formula sqrt(a²)=|a|.

Radikal belgisi ± ikkita qiymat beradi.

Radikal belgisi principal rootni beradi; ± tenglama yechimida qo‘llanadi.

Daraja qonunlari domain’dan mustaqil formal belgilar.

Rational exponents real sonlarda root mavjudligi va denominator restrictionsga bog‘liq.

sqrt(a+b)=sqrt(a)+sqrt(b).

Root sumga tarqalmaydi.

Maxrajni ratsionallashtirish faqat “eski usul”, matematik ahamiyati yo‘q.

U equivalent algebraic form, conjugate/factorization va keyingi symbolic manipulationsda foydali.

Turli radikallarni faqat calculator bilan taqqoslash mumkin.

Musbat radicals common integer power orqali exact taqqoslanadi.

Infinite radical uchun fixed-point equation yechishning o‘zi kifoya.

Convergence va admissible sign/root ham tekshirilishi kerak.

Testda A belgilangan bo‘lsa A albatta canonical.

Marked option proof emas; source bank mustaqil QA qilinadi.

Amaliy qo'llanilishi

Geometriya

Length, area va scale formulalarida square/cube roots va fractional powers tabiiy paydo bo‘ladi.

Fizika

Power-law modellar, period/frequency, inverse-square va dimensional formulas rational exponentsdan foydalanadi.

Engineering

Material scaling, geometric means va root-based tolerance formulalarini symbolic soddalashtirishda ishlatiladi.

Data science

Geometric mean, root transforms va power transformations fractional exponents bilan ifodalanadi.

Finance

Compound-growth formulasni t perioddan bir periodga qaytarishda t-th root yoki 1/t exponent ishlatiladi.

Computer graphics

Gamma/power transforms va nonlinear scalinglarda fractional exponents ishlatiladi.

Algebraic modelling

Irratsional tenglama/tengsizliklarni yechishdan oldin radicalsni exponent ko‘rinishiga o‘tkazish strukturani ochadi.

Test-bank QA

Parity, principal-root sign, exact exponent arithmetic va domain invariants OCR/answer xatolarini aniqlaydi.

Ratsional daraja va radikal: domain-first xarita

Ratsional ko‘rsatkich — universal workflow1. DOMAIN + PARITYm/n ni qisqartir • denominator juft/toq • maxraj ≠0principal even root ≥0EXPONENT ENGINEproduct → add exponentsquotient → subtractpower → multiplynegative → reciprocalRADICAL ENGINEperfect powers outeven root → |a|odd root keeps signlike radicals combineSTRUCTURE ENGINEcompare via common powerrationalize denominatorconjugate / cubesnested → rational exponentFINAL QA GATEreal domain • |a| on even roots • original denominator • exact exponent arithmeticinvalid even-root radicand → stop • marked option ≠ proof

Rational exponent/radical masalalarida domain va paritydan boshlanib exponent, radical yoki structural engine tanlanadi; yakunda principal-root sign va original-domain QA qilinadi.

Xulosa

Cheat sheet: a^(1/n)=sqrt[n](a); a^(m/n)=sqrt[n](a^m)=(sqrt[n](a))^m. a^(-r)=1/a^r (a≠0). a^r a^s=a^(r+s), a^r/a^s=a^(r-s), (a^r)^s=a^(rs) — real sohada domain shartlari bilan. Juft n: sqrt[n](a^n)=|a|; toq n: sqrt[n](a^n)=a. Juft indeksli ildiz uchun radikand≥0. Radical denominatorni one-term multiplier yoki conjugate bilan ratsionallashtiring. Har transformatsiyada original domainni saqlang.

Keyingi “Irratsional tenglamalar” mavzusida radikal ifoda tenglama ichida noma’lumga bog‘lanadi. U yerda eng muhim yangi qadam: domainni yozish, ildizlarni izolyatsiya qilish, darajaga ko‘tarish va extraneous rootsni original tenglamada tekshirish.

Bog'liq mavzular

Oldin bilishingiz kerak: Daraja va uning xossalari, darajali ifodalar, Ko'phadlar va ular ustida amallar, Qisqa ko'paytirish formulalari

Bog'liq mavzular: Modulli ifodalar va tenglamalar, Logarifmlar: hisoblashga doir masalalar

Keyingi mavzular: Irratsional tenglamalar, Irratsional tengsizliklar, Ko'rsatkichli tenglama

Manbalar

Ro'yxatdan o'tib, mashq qilishni boshlang