Câu b :
Ta có :
\(a+b+c=abc\)
\(\Leftrightarrow1=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\)
\(\Leftrightarrow2=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\Leftrightarrow4=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\) \(+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)\)
\(\)\(\Rightarrow\) \(\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+2=4\)
\(\Rightarrow\) \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\left(đpcm\right)\)