Đáp án:

$a)\sqrt{6}\\ b)1\\ c)16\\ d)-\sqrt{2}\\ e)\sqrt{2}\\ g)\sqrt{10}\\ h)0$

Giải thích các bước giải:

$a)\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\\ =\sqrt{9-2.3.\sqrt{6}+6}+\sqrt{9-2.3.2\sqrt{6}+24}\\ =\sqrt{(3-\sqrt{6})^2}+\sqrt{(3-2\sqrt{6})^2}\\ =|3-\sqrt{6}|+|3-2\sqrt{6}|\\ =3-\sqrt{6}+2\sqrt{6}-3\\ =\sqrt{6}\\ b)\sqrt{(2-\sqrt{3})^2}+\sqrt{4-2\sqrt{3}}\\ =|2-\sqrt{3}|+\sqrt{3-2.\sqrt{3}.1+1}\\ =2-\sqrt{3}+\sqrt{(\sqrt{3}-1)^2}\\ =2-\sqrt{3}+|\sqrt{3}-1|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\\ c)(\sqrt{10}+\sqrt{2})(6-2\sqrt{5})\sqrt{3+\sqrt{5}}\\ =(\sqrt{10}+\sqrt{2})(5-2\sqrt{5}+1)\sqrt{3+\sqrt{5}}\\ =(\sqrt{5}+1)(\sqrt{5}-1)^2.\sqrt{2}.\sqrt{3+\sqrt{5}}\\ =(\sqrt{5}+1)(\sqrt{5}-1)(\sqrt{5}-1)\sqrt{6+2\sqrt{5}}\\ =(5-1)(\sqrt{5}-1)\sqrt{5+2\sqrt{5}+1}\\ =4(\sqrt{5}-1)\sqrt{(\sqrt{5}+1)^2}\\ =4(\sqrt{5}-1)(\sqrt{5}+1)\\ =4(5-1)\\ =16\\ d)\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\\ =\dfrac{\sqrt{2}\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)}{\sqrt{2}}\\ =\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\\ =\dfrac{\sqrt{3-2\sqrt{3}+1}-\sqrt{3+2\sqrt{3}+1}}{\sqrt{2}}\\ =\dfrac{\sqrt{(\sqrt{3}-1)^2}-\sqrt{(\sqrt{3}+1)^2}}{\sqrt{2}}\\ =\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}\\ =\dfrac{-2}{\sqrt{2}}\\ =-\sqrt{2}\\ e)\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\\ =\dfrac{\sqrt{2}\left(\sqrt{3+\sqrt{5}}\right)}{\sqrt{2}}+\dfrac{\sqrt{2}\left(\sqrt{7-3\sqrt{5}}\right)}{\sqrt{2}}-\sqrt{2}\\ =\dfrac{\sqrt{6+2\sqrt{5}}}{\sqrt{2}}+\dfrac{\sqrt{14-6\sqrt{5}}}{\sqrt{2}}-\sqrt{2}\\ =\dfrac{\sqrt{5+2\sqrt{5}+1}}{\sqrt{2}}+\dfrac{\sqrt{9-2.3.\sqrt{5}+5}}{\sqrt{2}}-\sqrt{2}\\ =\dfrac{\sqrt{(\sqrt{5}+1)^2}}{\sqrt{2}}+\dfrac{\sqrt{(3-\sqrt{5})^2}}{\sqrt{2}}-\sqrt{2}\\ =\dfrac{\sqrt{5}+1}{\sqrt{2}}+\dfrac{3-\sqrt{5}}{\sqrt{2}}-\dfrac{2}{\sqrt{2}}\\ =\dfrac{2}{\sqrt{2}}\\ =\sqrt{2}\\ g)\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\\ =\dfrac{\sqrt{2}\left(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\right)}{\sqrt{2}}\\ =\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}}{\sqrt{2}}\\ =\dfrac{\sqrt{5+2\sqrt{5}+1}+\sqrt{5-2\sqrt{5}+1}}{\sqrt{2}}\\ =\dfrac{\sqrt{(\sqrt{5}+1)^2}+\sqrt{(\sqrt{5}-1)^2}}{\sqrt{2}}\\ =\dfrac{\sqrt{5}+1+\sqrt{5}-1}{\sqrt{2}}\\ =\dfrac{2\sqrt{5}}{\sqrt{2}}\\ =\sqrt{2}\sqrt{5}\\ =\sqrt{10}\\ h)\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\\ =\dfrac{\sqrt{2}\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\right)}{\sqrt{2}}\\ =\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2}{\sqrt{2}}\\ =\dfrac{\sqrt{7+2\sqrt{7}+1}-\sqrt{7-2\sqrt{7}+1}-2}{\sqrt{2}}\\ =\dfrac{\sqrt{(\sqrt{7}+1)^2}-\sqrt{(\sqrt{7}-1)^2}-2}{\sqrt{2}}\\ =\dfrac{\sqrt{7}+1-\sqrt{7}+1-2}{\sqrt{2}}\\ =0$