ta có : x\(^2\)\(\ge\)0
|x-1|\(\ge\)0
\(\Rightarrow\)x\(^2\)+|x-1|\(\ge\)0
\(\Rightarrow\)|x\(^2\)|x-1||=x\(^2\)|x-1|
\(\Rightarrow\)x\(^2\)+|x-1|=x\(^2\)+2
\(\Rightarrow\)|x-1|=2
TH1: x-1=2
\(\Rightarrow\)x=3
TH2: x-1=-2
\(\Rightarrow\)x=-1
vậy x\(\in\)\(\left\{3;-1\right\}\)