ĐXKĐ: \(\frac{-1+\sqrt{5}}{2}\le x\le\frac{1+\sqrt{5}}{2}\)
Áp dụng BĐT Cô-si:
\(x^2-x+2=\sqrt{x^2+x-1}+\sqrt{x-x^2+1}\le\frac{1+x^2+x-1}{2}+\frac{1+x-x^2+1}{2}\)
\(=\frac{2x+2}{2}=x+1\)
\(\Leftrightarrow x^2-x+2\le x+1\)
\(\Leftrightarrow x^2-2x+1\le0\)
\(\Leftrightarrow\left(x-1\right)^2\le0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x=1\) \(\left(TM\right)\)