a: |x-2|=x+1
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(x-2-x-1\right)\left(x-2+x+1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(2x-1\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{2}\)
b: |x+1|=x-2
\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(x-2-x-1\right)\left(x-2+x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\2x-1=0\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{2}\left(loại\right)\)
c: =>|2x-2|=x+2
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\\left(2x-2-x-2\right)\left(2x-2+x+2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\\left(3x-4\right)\cdot3x=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;\dfrac{4}{3}\right\}\)
d: |x-2|=2x-1
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\\left(2x-1-x+2\right)\left(2x-1+x-2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\\left(x+1\right)\left(3x-3\right)=0\end{matrix}\right.\Leftrightarrow x=1\)
