a: |1-x|=2x+1
=>|x-1|=2x+1
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+1\right)^2-\left(x-1\right)^2=0\\x>=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(2x+1+x-1\right)\left(2x+1-x+1\right)=0\\x>=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x\left(x+2\right)=0\\x>=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x=0\)
b: \(\left|-5x\right|=4-x\)
=>|5x|=4-x
\(\Leftrightarrow\left\{{}\begin{matrix}\left(5x\right)^2-\left(4-x\right)^2=0\\x< =4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(5x-4+x\right)\left(5x+4-x\right)=0\\x< =4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(6x-4\right)\left(4x+4\right)=0\\x< =4\end{matrix}\right.\Leftrightarrow x\in\left\{\dfrac{2}{3};-1\right\}\)
d: |3-x|=x-3
=>3-x<=0
=>x>=3
