Cho a, b, c là những số nguyên thỏa mãn: (1/a+1/b+1/c)^2=1/a^2+1/b^2+!/c^2 

Trả lời

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

$\iff {\left[\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{1}{c}\right]}^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$

Phân tích vế trái ta được: 

$\begin{array}{l}{\left(\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{1}{c}\right)}^2={\left(\frac{1}{a}+\frac{1}{b}\right)}^2+\frac{2}{ac}+\frac{2}{bc}+{\left(\frac{1}{c}\right)}^2=\frac{1}{a^2}+\frac{2}{ab}+\frac{1}{b^2}+\frac{2}{ac}+\frac{2}{bc}+\frac{1}{c^2}\\ \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{ac}+\frac{2}{bc}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\\ \Rightarrow 2.\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)=0\\ \Rightarrow \frac{c}{abc}+\frac{b}{abc}+\frac{a}{abc}=0\\ \Rightarrow a+b+c=0.abc=0\\ \Rightarrow a+b=-c\\ \Rightarrow -\left(a+b\right)=c\end{array}$