Question

Cho a, b, c thoa man \(\dfrac{1}{a}\)+\(\dfrac{1}{b}\)+\(\dfrac{1}{c}\)=2 va a+b+c=abc. Tinh M=\(\dfrac{1}{a}\)^2+\(\dfrac{1}{b}\)^2+\(\dfrac{1}{c}\)^2

Answer 1

ta có

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

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=4\)

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

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

\(\Leftrightarrow M=2\)