Bài 1: Ta có: \(\left|x+1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
Vậy \(x=2\) hoặc \(x=-4\)
Bài 2: Ta có: \(\left|x+1\right|\le3\)
Mà: \(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left|x+1\right|\le3\end{matrix}\right.\) Nên \(0\le\left|x+1\right|\le3\)
\(\Rightarrow\left|x+1\right|\in\left\{0;1;2;3\right\}\)
Ta có 4 trường hợp:
Trường hợp 1: Nếu \(\left|x+1\right|=0\)
\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Trường hợp 2: Nếu \(\left|x+1\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Trường hợp 3: Nếu \(\left|x+1\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
Trường hợp 4: Nếu \(\left|x+1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
Vậy \(x\in\left\{0;\pm1;\pm2;\pm3;-4\right\}\)
