Ta có:
\(2x+y-2\sqrt{xy}-6\sqrt{x}+9=0\) ĐK: x≥0; y≥0
⇔\(\left(x-2\sqrt{xy}+y\right)+\left(x-6\sqrt{x}+9\right)\)= 0
⇔\(\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{x}-3\right)^2=0\)
⇔\(\left(\sqrt{x}-\sqrt{y}\right)^2=0\) hoặc \(\left(\sqrt{x}-3\right)^2=0\)
⇔\(\sqrt{x}=\sqrt{y}\) hoặc \(\sqrt{x}=3\)
⇔x=y=9 (thỏa mãn)
Vậy x=y=9