Question

Cho a,b,c\(\ne\)0 thỏa a+b+c=0 thì

\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)

Answer 1

Ta có: \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)

\(=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}\)

Ta cần chứng minh: \(\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=0\) thật vậy:

\(\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=\dfrac{2\left(a+b+c\right)}{abc}=\dfrac{2.0}{abc}=0\)Tức là:\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\left(đpcm\right)\)