Question

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

Answer 1

+ \(a+b+c=abc\)

\(\Rightarrow\dfrac{a+b+c}{abc}=1\)

\(\Rightarrow\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{ab}=1\)

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

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\)

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

\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\)