Xét \(\sqrt{y\left(x-1\right)}+\sqrt{x^2-y}=x\sqrt{x}\) có:\(\sqrt{xy-y}-\sqrt{x^2-y}=\dfrac{xy-y-\left(x^2-y\right)}{\sqrt{y\left(x-1\right)}+\sqrt{x^2-y}}=\dfrac{x\left(y-x\right)}{x\sqrt{x}}=\dfrac{y-x}{\sqrt{x}}\)
\(\Rightarrow2\sqrt{xy-y}=\dfrac{y-x}{\sqrt{x}}+x\sqrt{x}=\dfrac{x^2-x+y}{\sqrt{x}}\)
\(\Rightarrow2\sqrt{y\left(x^2-x\right)}=x^2-x+y\)
\(\Rightarrow4y\left(x^2-x\right)=\left(x^2-x+y\right)^2\)
\(\Leftrightarrow\left(y-x^2+x\right)^2=0\Leftrightarrow y=x^2-x\). Thay vào pt(1) thì:
\(\sqrt{x^2+x-1}+\sqrt{-x^2+x+1}=x^2-x+2\)
Áp dụng BĐT AM-GM ta có:
\(\sqrt{x^2+x-1}\le\dfrac{x^2+x-1+1}{2}=\dfrac{x^2+x}{2}\)
\(\sqrt{-x^2+x+1}\le\dfrac{-x^2+x+1+1}{2}=\dfrac{-x^2+x+2}{2}\)
Cộng theo vế 2 BĐT trên ta có:
\(x^2-x+2\le\dfrac{x^2+x}{2}+\dfrac{-x^2+x+2}{2}=x+1\)
\(\Leftrightarrow x^2-2x+1\le0\Leftrightarrow\left(x-1\right)^2\le0\Rightarrow x=1\Rightarrow y=0\)
