|x+1|-|2-x|=0
TH1: x<-1
=> -x-1-2+x=0<=>0x=3(vo nghiem)
TH2 :-1\(\le x\le2\)
=>x+1-2+x=0<=>2x=1<=>x=\(\dfrac{1}{2}\) (thỏa mãn)
TH3 : x>2
=>x+1+2-x=0<=>0x=-3(vo nghiem)
Vậy S ={\(\dfrac{1}{2}\)}
\(\left|x+1\right|-\left|2-x\right|=0\\ \Leftrightarrow\left|x+1\right|=\left|2-x\right|\\ \Leftrightarrow\left[{}\begin{matrix}x+1=2-x\\x+1=x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+x=2-1\\x-x=-2-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=1\\0x=-3\left(Vô\text{ nghiệm }\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{2}\)
Vậy phương trình có nghiệm \(x=\dfrac{1}{2}\)
