Ta có: $\left \{ {{2.\sqrt{x-3}+\dfrac{12}{y-2x}=8} \atop {3.\sqrt{4x-12}+\dfrac{3}{2x-y}=\dfrac{9}{2}}} \right.$
⇔ $\left \{ {{2.\sqrt{x-3}-\dfrac{12}{2x-y}=8} \atop {3.\sqrt{4(x-3)}+\dfrac{3}{2x-y}=\dfrac{9}{2}}} \right.$
⇔ $\left \{ {{2.\sqrt{x-3}-\dfrac{12}{2x-y}=8} \atop {6.\sqrt{x-3}+\dfrac{3}{2x-y}=4,5}} \right.$
Đặt $\sqrt{x-3}=a$ $(a>0)$
Đặt $\dfrac{1}{2x-y}=b$ $(2x\neqy)$
Ta có hpt mới: $\left \{ {{2a-12b=8} \atop {6a+3b=4,5}} \right.$
⇔ $\left \{ {{a=1} \atop {b=-0,5}} \right.(tm)$
Khi $a=1$⇒ $\sqrt{x-3}=1$
⇔ $x=4$
$b=-0,5$⇒ $\dfrac{1}{2x-y}=-0,5$
⇒ $2x-y=-2$
⇔ $2.4-y=-2$
⇔ $y=10$
Vậy $(x;y)=(4;10)$
