Ta có :
\(\left|x^2+\left|x-1\right|\right|\ge0\)
\(\Rightarrow x^2+2\ge0\)
Mà 2 > 0
\(\Rightarrow x^2\ge0\)
Nên |x2 + |x - 1|| = x2 + 2
=> x2 + |x - 1| = x2 + 2
=> |x - 1| = 2
\(\Rightarrow\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)