Question

cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

Tính giá trị biểu thức : \(P=\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ca}{b^2}\)

Answer 1

Với \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\) 

Chứng minh : với \(a+b+c=0\) thì \(a=-\left(b+c\right)\Leftrightarrow a^3=-\left(b+c\right)^3\)

\(\Leftrightarrow a^3=-\left(b^3+c^3+3b^2c+3bc^2\right)\Leftrightarrow a^3+b^3+c^3=-\left(b^3+c^3+3b^2c+3bc^2\right)+b^3+c^3\)

\(\Leftrightarrow a^3+b^3+c^3=-3bc\left(b+c\right)=-3bc\left(-a\right)=3abc\)vì \(b+c=-a\) =>đpcm

Vì \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\)\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

Vậy \(P=\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ca}{b^2}=abc\left(\frac{1}{c^3}+\frac{1}{a^3}+\frac{1}{b^3}\right)=abc\frac{3}{abc}=3\)

**** mình nha