Đáp án:

$a)\sqrt{2}-1\\ b)\dfrac{\sqrt{5}}{2}\\ c)\sqrt{3}+\dfrac{3}{2}\\ d)\dfrac{\sqrt{5}+\sqrt{2}}{3}\\ e)(\sqrt{2}-1)(\sqrt{5}-2)\\ g)\dfrac{\sqrt{2}+2-\sqrt 6}{4}\\ h)\dfrac{6+3\sqrt{6}-3\sqrt{10}}{6}\\ i)\dfrac{9-6\sqrt{2}+5\sqrt{3}-3\sqrt{6}}{2}$

Giải thích các bước giải:

$a)\dfrac{3-2\sqrt{2}}{\sqrt{2}-1}\\ =\dfrac{(3-2\sqrt{2})(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}\\ =\dfrac{3\sqrt{2}+3-4-2\sqrt{2}}{2-1}\\ =\sqrt{2}-1\\ b)\dfrac{5-\sqrt{5}}{\sqrt{20}-2}\\ =\dfrac{(5-\sqrt{5})(\sqrt{20}+2)}{(\sqrt{20}-2)(\sqrt{20}+2)}\\ =\dfrac{5\sqrt{20}+10-\sqrt{100}-2\sqrt{5}}{20-4}\\ =\dfrac{5.2\sqrt{5}-2\sqrt{5}}{16}\\ =\dfrac{8\sqrt{5}}{16}\\ =\dfrac{\sqrt{5}}{2}\\ c)\dfrac{3}{4\sqrt{3}-6}\\ =\dfrac{3(4\sqrt{3}+6)}{(4\sqrt{3}-6)(4\sqrt{3}+6)}\\ =\dfrac{12\sqrt{3}+18}{48-36}\\ =\dfrac{12\sqrt{3}+18}{12}\\ =\sqrt{3}+\dfrac{3}{2}\\ d)\dfrac{\sqrt{\sqrt{5}+\sqrt{2}}}{\sqrt{3\sqrt{5}-3\sqrt{2}}}\\ =\sqrt{\dfrac{\sqrt{5}+\sqrt{2}}{3\sqrt{5}-3\sqrt{2}}}\\ =\sqrt{\dfrac{(\sqrt{5}+\sqrt{2})(3\sqrt{5}+3\sqrt{2})}{(3\sqrt{5}-3\sqrt{2})(3\sqrt{5}+3\sqrt{2})}}\\ =\sqrt{\dfrac{15+3\sqrt{10}+3\sqrt{10}+6}{45-18}}\\ =\sqrt{\dfrac{21+6\sqrt{10}}{27}}\\ =\sqrt{\dfrac{7+2\sqrt{10}}{9}}\\ =\sqrt{\dfrac{5+2\sqrt{5}\sqrt{2}+2}{9}}\\ =\sqrt{\dfrac{(\sqrt{5}+\sqrt{2})^2}{3^2}}\\ =\dfrac{\sqrt{5}+\sqrt{2}}{3}\\ e)\dfrac{1}{2+\sqrt{5}+2\sqrt{2}+\sqrt{10}}\\ =\dfrac{1}{2+\sqrt{5}+\sqrt{2}(2+\sqrt{5})}\\ =\dfrac{1}{(\sqrt{2}+1)(2+\sqrt{5})}\\ =\dfrac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)(2+\sqrt{5})}\\ =\dfrac{\sqrt{2}-1}{(2-1)(2+\sqrt{5})}\\ =\dfrac{\sqrt{2}-1}{2+\sqrt{5}}\\ =\dfrac{(\sqrt{2}-1)(\sqrt{5}-2)}{(\sqrt{5}+2)(\sqrt{5}-2)}\\ =\dfrac{\sqrt{10}-2\sqrt{2}-\sqrt{5}+2}{5-4}\\ =\sqrt{10}-2\sqrt{2}-\sqrt{5}+2\\ =\sqrt{2}(\sqrt{5}-2)-(\sqrt{5}-2)\\ =(\sqrt{2}-1)(\sqrt{5}-2)\\ g)\dfrac{1}{1+\sqrt{2}+\sqrt{3}}\\ =\dfrac{1+\sqrt{2}-\sqrt{3}}{(1+\sqrt{2}+\sqrt{3})(1+\sqrt{2}-\sqrt{3})}\\ =\dfrac{1+\sqrt{2}-\sqrt{3}}{(1+\sqrt{2})^2-3}\\ =\dfrac{1+\sqrt{2}-\sqrt{3}}{1+2\sqrt{2}+2-3}\\ =\dfrac{1+\sqrt{2}-\sqrt{3}}{2\sqrt{2}}\\ =\dfrac{\sqrt{2}+2-\sqrt 6}{4}\\ h)\dfrac{2\sqrt{3}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\\ =\dfrac{2\sqrt{3}(\sqrt{2}+\sqrt{3}-\sqrt{5})}{(\sqrt{2}+\sqrt{3}+\sqrt{5})(\sqrt{2}+\sqrt{3}-\sqrt{5})}\\ =\dfrac{2\sqrt{3}(\sqrt{2}+\sqrt{3}-\sqrt{5})}{(\sqrt{2}+\sqrt{3})^2-5}\\ =\dfrac{2\sqrt{3}(\sqrt{2}+\sqrt{3}-\sqrt{5})}{2+2\sqrt{6}+3-5}\\ =\dfrac{2\sqrt{6}+2\sqrt{9}-2\sqrt{15}}{2\sqrt{6}}\\ =\dfrac{2\sqrt{6}+6-2\sqrt{15}}{2\sqrt{6}}\\ =\dfrac{12+6\sqrt{6}-2\sqrt{90}}{12}\\ =\dfrac{12+6\sqrt{6}-6\sqrt{10}}{12}\\ =\dfrac{6+3\sqrt{6}-3\sqrt{10}}{6}\\ i)\dfrac{\sqrt{6}}{3+\sqrt{2}-\sqrt{3}}\\ =\dfrac{\sqrt{6}(\sqrt{2}-\sqrt{3}-3)}{(\sqrt{2}-\sqrt{3}+3)(\sqrt{2}-\sqrt{3}-3)}\\ =\dfrac{\sqrt{12}-\sqrt{18}-3\sqrt{6}}{(\sqrt{2}-\sqrt{3})^2-9)}\\ =\dfrac{2\sqrt{3}-3\sqrt{2}-3\sqrt{6}}{2-2\sqrt{6}+3-9}\\ =\dfrac{2\sqrt{3}-3\sqrt{2}-3\sqrt{6}}{-2\sqrt{6}-4}\\ =-\dfrac{2\sqrt{3}-3\sqrt{2}-3\sqrt{6}}{2(\sqrt{6}+2)}\\ =-\dfrac{(2\sqrt{3}-3\sqrt{2}-3\sqrt{6})(\sqrt{6}-2)}{2(\sqrt{6}+2)(\sqrt{6}-2)}\\ =-\dfrac{-18+12\sqrt{2}-10\sqrt{3}+6\sqrt{6}}{2(6-4)}\\ =-\dfrac{-18+12\sqrt{2}-10\sqrt{3}+6\sqrt{6}}{4}\\ =\dfrac{9-6\sqrt{2}+5\sqrt{3}-3\sqrt{6}}{2}$

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