+TH1 : x < 1
=> \(\left|x-1\right|+\left|1-x\right|=4-x\)
\(\Leftrightarrow\left|x-1\right|+\left|x-1\right|=4-x\)
\(\Leftrightarrow\left(1-x\right)+\left(1-x\right)=4-x\) ( do x < 1 nên \(\left|x-1\right|=1-x\) )
\(\Leftrightarrow2-2x=4-x\)
\(\Leftrightarrow-x=2\Leftrightarrow x=-2\)
+ TH2 : \(x\ge1\)
\(\Rightarrow\left|x-1\right|+\left|x-1\right|=4-x\)
\(\Leftrightarrow x-1+x-1=4-x\)
\(\Leftrightarrow2x-2=4-x\)
\(\Leftrightarrow3x=6\Leftrightarrow x=2\)