b, \(\left(x+1\right)^2-3\left|x+1\right|+2=0\)
TH1: \(x\ge-1\)
\(pt\Leftrightarrow\left(x+1\right)^2-3\left(x+1\right)+2=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
TH2: \(x< 1\)
\(pt\Leftrightarrow\left(x+1\right)^2+3\left(x+1\right)+2=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(tm\right)\\x=-3\left(tm\right)\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm ...
