Question

Tìm x, biết \(\left|x+1\right|-\left|x^2-1\right|=0\)

Answer 1

Lời giải:

Ta có:
\(|x+1|-|x^2-1|=0\)

\(\Leftrightarrow |x+1|-|(x+1)(x-1)|=0\)

\(\Leftrightarrow |x+1|(1-|x-1|)=0\)

\(\Rightarrow\left[{}\begin{matrix}\left|x+1\right|=0\left(1\right)\\1-\left|x-1\right|=0\left(2\right)\end{matrix}\right.\)

Có \((1)\Rightarrow x+1=0\Leftrightarrow x=-1\)

Có \((2)\Leftrightarrow 1=|x-1|\Leftrightarrow \) \(\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

Vậy \(x\in\left\{-1;0;2\right\}\)