\(\sqrt{x^2+x-1}+\sqrt{x-x^2+1}=x^2-x+2\)
ĐK: \(\dfrac{\sqrt{5}-1}{2}\le x\le\dfrac{\sqrt{5}+1}{2}\)
\(pt\Leftrightarrow\sqrt{x^2+x-1}-1+\sqrt{x-x^2+1}-1=x^2-x\)
\(\Leftrightarrow\dfrac{x^2+x-2}{\sqrt{x^2+x-1}+1}+\dfrac{x-x^2}{\sqrt{x-x^2+1}+1}=x^2-x\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+2\right)}{\sqrt{x^2+x-1}+1}-\dfrac{x\left(x-1\right)}{\sqrt{x-x^2+1}+1}=x\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{x+2}{\sqrt{x^2+x-1}+1}-\dfrac{x}{\sqrt{x-x^2+1}+1}-x\right)=0\)
Dễ thấy: \(\dfrac{x+2}{\sqrt{x^2+x-1}+1}-\dfrac{x}{\sqrt{x-x^2+1}+1}-x=0\) vô nghiệm
\(\Leftrightarrow x=1\)
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