Question

10,cho a+b+c=\(\frac{1}{9}\)abc và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\) với a,b,c khác 0.CMR:\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{2}\)

Answer 1

\(\Rightarrow\frac{a+b+c}{abc}=\frac{1}{9}\Leftrightarrow\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=\frac{2}{9}\)

Lại có \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=1\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=1\)

Vậy 1/a^2+1/b^2+1/c^2=1-2/9=7/9 ( Sê đài )