$\left \{ {{x+\sqrt{5}.y=\sqrt{5}} \atop {\sqrt{3}.x-y=\sqrt{3}}} \right.⇔$ $\left \{ {{x+y\sqrt{5}=\sqrt{5}(1)} \atop {y=x\sqrt{3}-\sqrt{3}(2)}} \right.$
Thế $(2)$ vào $(1)$, ta đc:
$x+(x\sqrt{3}-\sqrt{3})\sqrt{5}=\sqrt{5}$
$⇔x+x\sqrt{15}-\sqrt{15}=\sqrt{5}$
$⇔x(1+\sqrt{15})=\sqrt{5}+\sqrt{15}$
$⇔x=\frac{\sqrt{5}+\sqrt{15}}{1+\sqrt{15}}≈1,254$
$⇒y≈1,254.\sqrt{3}-\sqrt{3}≈0,44$
Vậy ...................................
