Question

[Nguyễn Tấn Trung]

Giups với

Answer 1

[Lương Ngọc Anh]

Ta có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\cdot\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

      = \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\cdot\frac{a^2bc+abc^2+ab^2c}{a^2b^2c^2}\)(1)

Mà a+c+b=0(2)

Từ (1)(2)=>\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)      =  \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\cdot\frac{abc\left(a+b+c\right)}{a^2b^2c^2}\)= \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

=>  \(\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}\)(đpcm)

Answer 2

[Võ Đông Anh Tuấn]

Ta có : \(đt\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\) ( do cả hai vế đều dương )

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

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

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\) ( đúng đo \(a+b+c=0\) )