a/
0<|x+1| ≤ 3
⇒|x+1|∈{1;2;3}
⇔+)\(\left[ \begin{array}{l}x+1=1\\x+1=-1\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=0\\x=-2\end{array} \right.\)
+)\(\left[ \begin{array}{l}x+1=2\\x+1=-2\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=1\\x=-3\end{array} \right.\)
+)\(\left[ \begin{array}{l}x+1=3\\x+1=-3\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=2\\x=-4\end{array} \right.\)
b/
$-(5+x-3)-|4|=5-(7+4)$
$⇔-5-x+3-4=5-11$
$⇔-6-x=-6$
$⇔-x=-6+6$
$⇔-x=0$
$⇔x=0$
Vậy x=0
