Question

cho a+b+c=1 và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) . Chứng minh rằng : \(a^2+b^2+c^2\)=1

Answer 1

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+ca+bc=0\)

ta có \(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)=a^2+b^2+c^2=1\)

Answer 2

Ta có: \(a+b+c=1\Rightarrow\left(a+b+c\right)^2=1\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=1\left(1\right)\)

Ta lại có:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow ab+bc+ca=0\left(2\right)\)

Lấy (1) - (2), ta được:

\(a^2+b^2+c^2=1\left(đpcm\right)\)