Question

Cho (a+b+c)^2=a^2+b^2+c^2 và abc khác 0

CM \(\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}=3\)

Answer 1

\(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow2\left(ab+bc+ca\right)=0\)

\(\Leftrightarrow ab+bc+ca=0\Leftrightarrow\dfrac{ab+bc+ca}{abc}=0\)\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) (1)

Ta có: \(a+b+c=0\Rightarrow a^3+b^3+c^3=3abc\) (Bn thự cm nhé)

(1) \(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\Leftrightarrow abc\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=3\)

\(\Leftrightarrow\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}=3\left(đpcm\right)\)