Question

Cho ba số a, b, c khác 0 thỏa mãn: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\). Hãy tính \(P=\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ac}{b^2}\).

Answer 1

Lời giải:

Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow \frac{ab+bc+ac}{abc}=0\)

\(\Leftrightarrow ab+bc+ac=0\)

\(\Rightarrow ab+bc=-ac\). Khi đó:

\(P=\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ca}{b^2}=\frac{(ab)^3+(bc)^3+(ca)^3}{a^2b^2c^2}\)

\(=\frac{(ab+bc)^3-3(ab)^2bc-3ab(bc)^2+(ca)^3}{a^2b^2c^2}\)

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

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