Rút gọn biểu thức: a) √(4-3 √2) ² b) √(2+ √5) ² c) √(4+ √2) ² d) √6- 2 √5 e) √7+4 √3 f) √12-6 √3 g) √17+12 √2 h) √2- √11+6 √2 / √6+2 √5- √5 i)

$a)\sqrt{(4-3\sqrt{2})^2}=|4-3\sqrt{2}|=3\sqrt{2}-4\\b)\sqrt{(2+\sqrt{5})^2}=|2+\sqrt{5}|=2+\sqrt{5}\\c)\sqrt{(4+\sqrt{2})^2}=|4+\sqrt{2}|=4+\sqrt{2}\\d)\sqrt{6-2\sqrt{5}}=\sqrt{5-2\sqrt{5}+1}=\sqrt{(\sqrt{5}-1)^2}=|\sqrt{5}-1|=\sqrt{5}-1\\e)\sqrt{7+4\sqrt{3}}=\sqrt{4+2.2.\sqrt{3}+3}=\sqrt{(2+\sqrt{3})^2}=|2+\sqrt{3}|=2+\sqrt{3}\\f)\sqrt{12-6\sqrt{3}}=\sqrt{9-2.3.\sqrt{3}+3}=\sqrt{(3-\sqrt{3})^2}=|3-\sqrt{3}|=3-\sqrt{3}\\g)\sqrt{17+12\sqrt{2}}=\sqrt{9+2.3.2\sqrt{2}+8}=\sqrt{(3+2\sqrt{2})^2}=|3+2\sqrt{2}|=3+2\sqrt{2}$

$h)\dfrac{\sqrt{2}-\sqrt{11+6\sqrt{2}}}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}=\dfrac{\sqrt{2}-\sqrt{9+2.3.\sqrt{2}+2}}{\sqrt{1+2.1.\sqrt{5}+5}-\sqrt{5}}=\dfrac{\sqrt{2}-\sqrt{(3+\sqrt{2})^2}}{\sqrt{(1+\sqrt{5})^2}-\sqrt{5}}=\dfrac{\sqrt{2}-|3+\sqrt{2}|}{|1+\sqrt{5}|-\sqrt{5}}=\dfrac{\sqrt{2}-3-\sqrt{2}}{1+\sqrt{5}-\sqrt{5}}=\dfrac{-3}{1}=-3\\i)\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{3-2\sqrt{3}+1}}=\sqrt{6+2\sqrt{(\sqrt{3}-1)^2}}=\sqrt{6+2|\sqrt{3}-1|}=\sqrt{6+2\sqrt{3}-2}=\sqrt{3+2\sqrt{3}+1}=\sqrt{(\sqrt{3}+1)^2}=|\sqrt{3}+1|=\sqrt{3}+1$