Question

cho 1/a +1 /b +1/c = 0 . Tính

A= bc/a^2 + ca/b^2 +ab/c^2

Answer 1

Đặt \(x=\dfrac{1}{a},y=\dfrac{1}{b},z=\dfrac{1}{c}\Rightarrow x+y+z=0\)

=> x+y=-z

Vì\(\left(x+y+z\right)^3=0\Leftrightarrow\left(x+y\right)^3+z^3+3\left(x+y\right).z.\left(x+y+z\right)=0\Leftrightarrow\left(x+y\right)^3+z^3=0\)<=> \(x^3+y^3+z^3+3xy\left(x+y\right)=0\)

<=> \(x^3+y^3+z^3=-3xy\left(x+y\right)=3xyz\)

Khi đó A = \(\dfrac{x^2}{yz}+\dfrac{y^2}{xz}+\dfrac{z^2}{xy}==\dfrac{x^3+y^3+z^3}{xyz}=\dfrac{3xyz}{xyz}=3\)

Vậy A=3