Question

[JESSICA]

Cho 3 số thực a,b,c\(\ne0\)và \(\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\). Chứng minh:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

Answer 1

[Hoàng Thị Lan Hương]

ĐK \(\hept{\begin{cases}a,b,c\ne0\\a+b\ge0;a+c\ge0;b+c\ge0\end{cases}}\)

Ta có \(\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\Rightarrow a+b=a+b+2c+2\sqrt{\left(a+c\right)\left(b+c\right)}\)

\(\Leftrightarrow-c=\sqrt{\left(a+c\right)\left(b+c\right)}\Leftrightarrow\hept{\begin{cases}c\le0\\c^2=ab+ac+bc+c^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}c\le0\\ab+ac+bc=0\end{cases}\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ac+ab}{abc}}=\frac{0}{abc}=0\left(đpcm\right)\)

Vậy \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)