Question

\(|x^2+|x-1||=x^2+2\)

Answer 1

Vi x²+2≥0 nen pt⇔ x²+/x-1/=x²+2

=> /x-1/=2

=> \(\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=1+2=3\\x=-2+1=-1\end{matrix}\right.\)

Answer 2

\(\left|x^2+|x-1|\right|=x^2+2\)

Dễ nhận thấy: \(x^2+2>0\) (*)

Ta có: \( \left|x^2+|x-1|\right|=x^2+2\Leftrightarrow\left[{}\begin{matrix}x^2+ \left|x-1\right|=x^2+2\\x^2+\left|x-1\right|=-x^2-2\end{matrix}\right.\)

Từ (*) ta có th 2 loại

Nên: \(x^2+\left|x-1\right|=x^2+2\Leftrightarrow\left|x-1\right|=2\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)