\(\left|1-x\right|+x+3=0\)
+) TH1: \(1-x\ge0\Rightarrow x\le1\)
Ta có: \(1-x+x+3=0\)
\(\Rightarrow4=0\) (vô lí)
+) TH2: \(1-x< 0\Rightarrow x>1\)
Ta có: \(-1+x+x+3=0\)
\(\Rightarrow2x+2=0\)
\(\Rightarrow2x=-2\)
\(\Rightarrow x=-1\) (TM)
Vậy \(x=-1.\)