ĐK,x\(\ge1,y\ge1\)
Ta có \(\sqrt{x^2+5}+\sqrt{x-1}+x^2=\sqrt{y^2+5}+\sqrt{y-1}+y^2\Leftrightarrow\left(\sqrt{x^2+5}-\sqrt{y^2+5}\right)+\left(\sqrt{x-1}-\sqrt{y-1}\right)+\left(x^2-y^2\right)=0\Leftrightarrow\dfrac{x^2+5-\left(y^2+5\right)}{\sqrt{x^2+5}+\sqrt{y^2+5}}+\dfrac{x-1-\left(y-1\right)}{\sqrt{x-1}+\sqrt{y-1}}+\left(x-y\right)\left(x+y\right)=0\Leftrightarrow\dfrac{\left(x-y\right)\left(x+y\right)}{\sqrt{x^2+5}+\sqrt{y^2+5}}+\dfrac{x-y}{\sqrt{x-1}+\sqrt{y-1}}+\left(x-y\right)\left(x+y\right)=0\Leftrightarrow\left(x-y\right)\left(\dfrac{x+y}{\sqrt{x^2+5}+\sqrt{y^2+5}}+\dfrac{1}{\sqrt{x-1}+\sqrt{y-1}}+x+y\right)=0\)(*)
Ta lại có \(\dfrac{x+y}{\sqrt{x^2+5}+\sqrt{y^2+5}}+\dfrac{1}{\sqrt{x-1}+\sqrt{y-1}}+x+y>0\)
Vậy (*)\(\Leftrightarrow x-y=0\Leftrightarrow x=y\)
Vậy \(\sqrt{x^2+5}+\sqrt{x-1}+x^2=\sqrt{y^2+5}+\sqrt{y-1}+y^2\) thì x=y
