Question

Chứng minh rằng : Nếu \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\) và a + b + c = abc thì \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\)

Answer 1

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

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

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

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

=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2.\dfrac{a+b+c}{abc}\) = 4

=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\) + \(2.\dfrac{abc}{abc}\) = 4 ( vì a+b + c = abc)

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