Ta có:
\(\left|x\right|=x+1\Rightarrow\left[{}\begin{matrix}x=x+1\\x=-x-1\end{matrix}\right.\)
TH1: \(x=x+1\)
⇒Không có x thỏa mãn
TH2:\(x=-x-1\)
\(\Rightarrow1=-x-x=-2x\)
\(\Rightarrow x=\frac{-1}{2}\)
Vậy \(x=\frac{-1}{2}\) thỏa mãn đề bài.
\(\left|x\right|=x+1\)
\(\Rightarrow\left[{}\begin{matrix}x=x+1\\x=-x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-x=1\\x+x=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}0x=1\\2x=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x=\left(-1\right):2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x=-\frac{1}{2}\end{matrix}\right.\)
Vậy \(x=-\frac{1}{2}.\)
Chúc bạn học tốt!
