Question

[Akatsuki Hikari]

(a+b+c)²= a²+b²+c² và abc≠0. Chứng minh rằng \(\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}=3\)

Answer 1

[Ma Sói]

Ta có:

\(\left(a+b+c\right)^2=a^2+b^2+c^2\)

\(a^2+b^2+c^2+2ac+2ab+2bc=a^2+b^2+c^2\)

\(ab+bc+ca=0\)

\(ab+bc=-ac\)

\(\left(ab+bc\right)^3=-a^3c^3\)

\(a^3c^3+a^3b^3+b^3c^3+3ab^2c\left(ab+bc\right)=0\)

\(a^3c^3+a^3b^3+b^3c^3=-3ab^2c\left(-ac\right)\)

\(a^3c^3+a^3b^3+b^3c^3=3a^2b^2c^2\)

Ta có:

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