Đáp án:
$a)2\\ b)2$
Giải thích các bước giải:
$a)\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\dfrac{1}{2-\sqrt{3}}-\dfrac{6}{\sqrt{3}}\\ =\dfrac{(\sqrt{15}-\sqrt{12})(\sqrt{5}+2)}{(\sqrt{5}-2)(\sqrt{5}+2)}+\dfrac{2-\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})}-\dfrac{6\sqrt{3}}{3}\\ =\dfrac{\sqrt{75}+2\sqrt{15}-\sqrt{60}-2\sqrt{12}}{5-4}+\dfrac{2-\sqrt{3}}{4-3}-2\sqrt{3}\\ =\sqrt{75}+2\sqrt{15}-\sqrt{60}-2\sqrt{12}+2-\sqrt{3}-2\sqrt{3}\\ =5\sqrt{3}+2\sqrt{15}-2\sqrt{15}-4\sqrt{3}+2-\sqrt{3}-2\sqrt{3}\\ =2\\ b)\dfrac{1}{2}\sqrt{48}-3\sqrt{\dfrac{1}{3}}+\sqrt{7-4\sqrt{3}}\\ =\dfrac{1}{2}\sqrt{16.3}-\dfrac{3}{\sqrt{3}}+\sqrt{4-2.2\sqrt{3}+3}\\ =\dfrac{1}{2}.4\sqrt{3}-\sqrt{3}+\sqrt{4-2.2\sqrt{3}+3}\\ =2\sqrt{3}-\sqrt{3}+\sqrt{(2-\sqrt{3})^2}\\ =2\sqrt{3}-\sqrt{3}+2-\sqrt{3}\\ =2$