Sửa đề `x->6`
$=\dfrac{\sqrt{6}(3+\sqrt{2}+\sqrt{3})}{9+6\sqrt{2}+2-3}$ `(`Nhân lượng liên hợp`)`
$=\dfrac{\sqrt{6}(3+\sqrt{2}+\sqrt{3})}{8+6\sqrt{2}}$
$=\dfrac{\sqrt{3}(3+\sqrt{2}+\sqrt{3})}{4\sqrt{2}+6}$
$=\dfrac{(3\sqrt{3}+\sqrt{6}+3)(4\sqrt{2}-6)}{(4\sqrt{2})^2-6^2}$
$=\dfrac{12\sqrt{6}-18\sqrt{3}+8\sqrt{3}-6\sqrt{6}+12\sqrt{2}-18}{32-36}$
$=\dfrac{6\sqrt{6}-10\sqrt{3}+12\sqrt{2}-18}{-4}$
$=\dfrac{2(3\sqrt{6}-5\sqrt{3}+6\sqrt{2}-9)}{-4}$
$=\dfrac{-3\sqrt{6}+5\sqrt{3}-6\sqrt{2}+9}{2}$
