Đáp án:
$-\dfrac{\sqrt{6}}{2}$
Giải thích các bước giải:
$\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}$
$=\dfrac{2\sqrt{4}-\sqrt{6}}{\sqrt{9}-\sqrt{24}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}(\sqrt{5}+\sqrt{27})}$
$=\dfrac{4-\sqrt{6}}{3-2\sqrt{6}}-\dfrac{1}{\sqrt{6}}$
$=\dfrac{(4-\sqrt{6}).\sqrt{6}-3+2\sqrt{6}}{(3-2\sqrt{6}).\sqrt{6}}$
$=\dfrac{6\sqrt{6}-9}{3\sqrt{6}-12}$
$=\dfrac{2\sqrt{6}-3}{\sqrt{6}-4}$
$=\dfrac{2\sqrt{6}-8+5}{\sqrt{6}-4}$
$=2+\dfrac{5}{\sqrt{6}-4}$
$=2+\dfrac{5(\sqrt{6}+4)}{(\sqrt{6}-4)(\sqrt{6}+4)}$
$=2+\dfrac{5\sqrt{6}+20}{-10}$
$=\dfrac{-20+5\sqrt{6}+20}{-10}$
$=-\dfrac{\sqrt{6}}{2}$
