$y=\frac{k}{x+\sqrt{x}}$
แทนค่า $x=2,y=1$
$1=\frac{k}{2+\sqrt{2}}$---->$k=2+\sqrt{2}$
ดังนั้น $y=\frac{2+\sqrt{2}}{x+\sqrt{x}}$ แทนค่า $y=\frac{1}{2-\sqrt{2}}$
$\frac{1}{2-\sqrt{2}}=\frac{2+\sqrt{2}}{x+\sqrt{x}}$
$x+\sqrt{x}=(2+\sqrt{2})(2-\sqrt{2})=2$
$x+\sqrt{x}-2=0$
$(\sqrt{x}+2)(\sqrt{x}-1)=0$
$\therefore \sqrt{x}=1$
$x=1$