Question

Cho a,b,c khác 0 thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\) và \(a+b+c=abc\)

Tính \(P=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

Answer 1

Ta có:

\(a+b+c=abc\)

\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)

Ta lại có:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)

\(\Rightarrow P=2\)