Agar $a+b+c=12$ va $\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{2}$ bo'lsa, $\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}$ ning qiymatini toping.
- A. $3$
- B. $-\dfrac{3}{7}$
- C. $\dfrac{7}{3}$
- D. $6$
Algebraik kasrlar bo‘yicha savol
Javob va yechimni ko‘rish
Javob: $3$
Har bir suratni $c=12-(a+b)$ kabi yozamiz. Yig‘indi $12\cdot\frac12-3=3$.
Mavzuni mustahkamlang
Bu savol Algebraik kasrlar mavzusiga tegishli. Ta’riflar, formulalar va misollarni mavzu sahifasida ko‘ring.
Mavzuni o‘rganishShu mavzudagi boshqa savollar
Ifodani soddalashtiring: $\frac{x^2 - 9}{x^2 + 6x + 9} \cdot \frac{x + 3}{x - 3}$Kasrlarni ayiring: $\frac{2a}{a^2 - b^2} - \frac{1}{a - b}$Tenglamani yeching: $\frac{x}{x-2} - \frac{2}{x-2} = 1$Ifodani soddalashtiring: $\left(\frac{1}{x} + \frac{1}{y}\right) : \frac{x^2 - y^2}{xy}$Agar $x + \frac{1}{x} = 5$ bo'lsa, $x^2 + \frac{1}{x^2}$ ning qiymatini toping.