\[\huge 4)\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\\=\frac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\frac{\sqrt{5}+3\sqrt{3}}{\sqrt{30}+9\sqrt{2}}\\=\frac{(4\sqrt{2}-2\sqrt{3})(3\sqrt{2}+4\sqrt{3})}{-30}-\frac{(\sqrt{5}+3\sqrt{3})(\sqrt{30}-9\sqrt{2})}{-132}\\=\frac{2(2\sqrt{2}-\sqrt{3})(3\sqrt{2}+4\sqrt{3})}{-30}-\frac{\sqrt{150}-9\sqrt{10}+3\sqrt{90}-27\sqrt{6}}{-132}\\=\frac{(2\sqrt{2}-\sqrt{3})(3\sqrt{2}+4\sqrt{3})}{-15}-\frac{\sqrt{150}-9\sqrt{10}+9\sqrt{10}-27\sqrt{6}}{-132}\\=\frac{12+8\sqrt{6}-3\sqrt{6}-12}{-15}-\frac{5\sqrt{6}-27\sqrt{6}}{-132}\\=\frac{5\sqrt{6}}{-15}-\frac{-22\sqrt{6}}{-132}\\=\frac{\sqrt{6}}{-3}-\frac{\sqrt{6}}{6}\\=-\frac{\sqrt{6}}{3}-\frac{\sqrt{6}}{6}\\=-\frac{\sqrt{6}}{2}\]
em thử latex ạ nhưng đây cũng là bài giải của em