\(\left|\left|x+2\right|-3\right|=1\)
\(\Rightarrow\left[\begin{array}{nghiempt}\left|x+2\right|-3=1\\3-\left|x+2\right|=1\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}\left|x+2\right|=4\\\left|x+2\right|=2\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+2=4\\x+2=-4\\x+2=2\\x+2=-2\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=-6\\x=0\\x=-4\end{array}\right.\)
Vậy \(x\in\left\{-6;-4;2;0\right\}\)
|| x + 2 | - 3 | = 1
\(\Rightarrow\left|x+2\right|-3=\pm1\)
Xét \(\left|x+2\right|-3=1\Rightarrow\left|x+2\right|=4\)
\(\Rightarrow x+2=\pm4\)
+) \(x+2=4\Rightarrow x=2\)
+) \(x+2=-4\Rightarrow x=-6\)
Xét \(\left|x+2\right|-3=-1\Rightarrow\left|x+2\right|=2\)
\(\Rightarrow x+2=\pm2\)
+) \(x+2=2\Rightarrow x=0\)
+) \(x+2=-2\Rightarrow x=-4\)
Vậy \(x\in\left\{2;-6;0;-4\right\}\)
