ĐK:\(a\ge0;b\ge0;c\ge0;a-b+c\ge0\)
Ta có:
\(\sqrt{a-b+c}=\sqrt{a}-\sqrt{b}+\sqrt{c}\)
\(\Leftrightarrow\sqrt{a-b+c}+\sqrt{b}=\sqrt{a}+\sqrt{c}\)
\(\Leftrightarrow\left(a-b+c+b\right)+2\sqrt{\left(a-b+c\right)b}=\left(a+c\right)+2\sqrt{ac}\)
\(\Leftrightarrow\left(a+c\right)+2\sqrt{\left(a-b+c\right)b}=\left(a+c\right)+2\sqrt{ac}\)
\(\Leftrightarrow\sqrt{\left(a-b+c\right)b}=\sqrt{ac}\)
\(\Leftrightarrow\left(a-b+c\right)b=ac\)
\(\Leftrightarrow ab-b^2+bc-ac=0\)
\(\Leftrightarrow b\left(a-b\right)-c\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(b-c\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\end{matrix}\right.\)
Vậy có hai trường hợp:
a=b hoặc b=c thì \(\sqrt{a-b+c}=\sqrt{a}-\sqrt{b}+\sqrt{c}\)
