Đáp án:

$a)\dfrac{\sqrt{3}-1}{\sqrt{2}}\\ b)\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\\ c)\dfrac{\sqrt{7}-\sqrt{3}}{\sqrt{2}}\\ f)\sqrt{2}+\sqrt{3}+\sqrt{5}\\ g)1+\sqrt{2}+\sqrt{5}$

Giải thích các bước giải:

$a)\sqrt{2-\sqrt{3}}\\ =\dfrac{\sqrt{2}\sqrt{2-\sqrt{3}}}{\sqrt{2}}\\ =\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\\ =\dfrac{\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}\\ =\dfrac{\sqrt{(\sqrt{3}-1)^2}}{\sqrt{2}}\\ =\dfrac{\sqrt{3}-1}{\sqrt{2}}\\ b)\sqrt{4+\sqrt{15}}\\ =\dfrac{\sqrt{2}\sqrt{4+\sqrt{15}}}{\sqrt{2}}\\ =\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\\ =\dfrac{\sqrt{5+2\sqrt{5}.\sqrt{3}+3}}{\sqrt{2}}\\ =\dfrac{\sqrt{(\sqrt{5}+\sqrt{3})^2}}{\sqrt{2}}\\ =\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\\ c)\sqrt{5-\sqrt{21}}\\ =\dfrac{\sqrt{2}\sqrt{5-\sqrt{21}}}{\sqrt{2}}\\ =\dfrac{\sqrt{10-2\sqrt{21}}}{\sqrt{2}}\\ =\dfrac{\sqrt{7-2\sqrt{7}.\sqrt{3}+3}}{\sqrt{2}}\\ =\dfrac{\sqrt{(\sqrt{7}-\sqrt{3})^2}}{\sqrt{2}}\\ =\dfrac{\sqrt{7}-\sqrt{3}}{\sqrt{2}}\\ d)\sqrt{38-12\sqrt{5}}\\ e)\sqrt{8-\sqrt{35}}\\ f)\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\\ =\sqrt{2+2\sqrt{6}+3+5+2\sqrt{10}+2\sqrt{15}}\\ =\sqrt{2+2\sqrt{2}.\sqrt{3}+3+5+2\sqrt{10}+2\sqrt{15}}\\ =\sqrt{(\sqrt{2}+\sqrt{3})^2+5+2\sqrt{10}+2\sqrt{15}}\\ =\sqrt{(\sqrt{2}+\sqrt{3})^2+2(\sqrt{2}+\sqrt{3})\sqrt{5}+5}\\ =\sqrt{(\sqrt{2}+\sqrt{3}+\sqrt{5})^2}\\ =\sqrt{2}+\sqrt{3}+\sqrt{5}\\ g)\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}\\ =\sqrt{8+2\sqrt{2}+\sqrt{20}+\sqrt{40}}\\ =\sqrt{1+2\sqrt{2}+2+5+2\sqrt{5}+2\sqrt{10}}\\ =\sqrt{(1+\sqrt{2})^2+5+2\sqrt{5}+2\sqrt{10}}\\ =\sqrt{(1+\sqrt{2})^2+2(1+\sqrt{2})\sqrt{5}+5}\\ =\sqrt{(1+\sqrt{2}+\sqrt{5})^2}\\ =1+\sqrt{2}+\sqrt{5}$