Bài 1:
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\) bình phương hai vế ta có:
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=2^2\)
=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)
=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\frac{\left(a+b+c\right)}{a.b.c}=4\) (Quy đồng \(MTC=a.b.c\))
=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2.a.b.c}{a.b.c}=4\) (Vì \(a+b+c=a.b.c\))
=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)
=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2\)
=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\left(đpcm\right).\)
Chúc bạn học tốt!
