a) Ta có: |-2x+3|-4+x=0(1)
Trường hợp 1: \(x\le\frac{3}{2}\)
(1)\(\Leftrightarrow-2x+3-4+x=0\)
\(\Leftrightarrow-x-1=0\)
\(\Leftrightarrow-x=1\)
hay x=-1(nhận)
Trường hợp 2: \(x>\frac{3}{2}\)
(1)\(\Leftrightarrow2x-3-4+x=0\)
\(\Leftrightarrow3x-7=0\)
\(\Leftrightarrow3x=7\)
hay \(x=\frac{7}{3}\)(loại)
Vậy: \(x\in\left\{-1;\frac{7}{3}\right\}\)
b) Ta có: \(\left|\frac{1}{2}x+2\right|-x+5=0\)(2)
Trường hợp 1: \(x\ge-4\)
(2)\(\Leftrightarrow\frac{1}{2}x+2-x+5=0\)
\(\Leftrightarrow\frac{-1}{2}x+7=0\)
\(\Leftrightarrow\frac{-1}{2}x=-7\)
hay \(x=-7:\frac{-1}{2}=-7\cdot\left(-2\right)=14\)(nhận)
Trường hợp 2: x<-4
(2)\(\Leftrightarrow\frac{-1}{2}x-2-x+5=0\)
\(\Leftrightarrow\frac{-3}{2}x+3=0\)
\(\Leftrightarrow\frac{-3}{2}x=-3\)
hay \(x=-3:\frac{-3}{2}=-3\cdot\frac{2}{-3}=2\)(loại)
Vậy: x=14
