[tex]\sqrt{x^{2}+x-1}+\sqrt{x-x^{2}+1}=x^2-x+2[/tex]
Ta có [tex]\sqrt{(x^{2}+x-1)1}\leq \frac{x^{2}+x-1+1}{2}= \frac{x^{2}+x}{2}[/tex]
[tex]\sqrt{(x-x^{2}+1)1}\leq \frac{x-x^{2}+2}{2} \rightarrow VT\leq x+1 (x-1)^{2}\geq 0\rightarrow x^{2}-x+2\geq x+1\rightarrow VP\geq VT\rightarrow x=1[/tex]