Question

Rút gọn biểu thức R = \(\sqrt{x^2-2x+1}+\sqrt{x^2+2x+1}\)

biết \(-1\le x\le1\)

Answer 1

Ta có: \(R=\sqrt{x^2-2x+1}+\sqrt{x^2+2x+1}\)

\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+1\right)^2}\)

\(=\left|x-1\right|+\left|x+1\right|\)

Ta có: \(-1\le x\le1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\le1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\x-1\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left|x+1\right|=x+1\\\left|x-1\right|=1-x\end{matrix}\right.\)

\(\Leftrightarrow R=x+1+1-x=2\)