$\left \{ {{x+2\sqrt{xy} + y = 25} \atop {\sqrt{x}-\sqrt{y}=1}} \right.$
ĐKXĐ: $x,y ≥ 0$
$⇔ \left \{ {{(\sqrt{x}+\sqrt{y})² = 25} \atop {\sqrt{x}-\sqrt{y}=1}} \right.$
$⇔ \left \{ {{\sqrt{x}+\sqrt{y} = 5} \atop {\sqrt{x}-\sqrt{y}=1}} \right.$ (Vì $\sqrt{x}+\sqrt{y} ≥ 0$ )
$⇔ \left \{ {{\sqrt{x}+\sqrt{y} +\sqrt{x}-\sqrt{y}=6 } \atop {\sqrt{x}-\sqrt{y}=1}} \right.$
$⇔ \left \{ {{\sqrt{x}=3} \atop {\sqrt{x}-\sqrt{y}=1}} \right.$
$⇔ \left \{ {{\sqrt{x}=3} \atop {\sqrt{y}=2}} \right.$
$⇔ \left \{ {{x=9} \atop {y=4}} \right.$ (TM)
Vậy $x = 9 , y = 4$
