Question

CMR : 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:

\(\dfrac{1}{a^2}+\dfrac{1}{ab}+\dfrac{1}{ac}=\dfrac{2}{a}\)

\(\dfrac{1}{ab}+\dfrac{1}{b^2}+\dfrac{1}{bc}=\dfrac{2}{b}\)

\(\dfrac{1}{ac}+\dfrac{1}{bc}+\dfrac{1}{c^2}=\dfrac{2}{c}\)

Cộng vế với vế ta được:

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

\(\Leftrightarrow\)\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2.\dfrac{c+a+b}{abc}=2.2\)

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

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

Answer 2

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

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

<=>\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{ac}+\dfrac{2}{bc}=4\) (hằng đẳng thức thứ 8)

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

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

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

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