Đáp án:
\(1/\\ a,\ 4\sqrt{3}\\ b,\ 8\\ c,\ -1\\ 2/\\ 2/\\ a,\ -4\sqrt{2}\\ b,\ \dfrac{\sqrt{6}}{2}\)
Giải thích các bước giải:
\(1/\\ a,\ \dfrac{2}{3}\sqrt{27}-6\sqrt{\dfrac{1}{3}}+\sqrt{48}\\ =\sqrt{\dfrac{4}{9}.27}-\sqrt{36.\dfrac{1}{3}}+\sqrt{48}\\ =\sqrt{12}-\sqrt{12}+\sqrt{48}\\ =(\sqrt{12}-\sqrt{12})+4\sqrt{3}\\ =4\sqrt{3}\\ b,\ \bigg(\sqrt{\dfrac{9}{2}}+\sqrt{32}-3\sqrt{\dfrac{1}{2}}\bigg).\sqrt{2}\\ =\sqrt{\dfrac{9}{2}.2}+\sqrt{32.2}-\sqrt{9.\dfrac{1}{2}.2}\\ =\sqrt{9}+\sqrt{64}-\sqrt{9}\\ =(\sqrt{9}-\sqrt{9})+\sqrt{64}\\ =\sqrt{64}=|8|=8\\ c,\ \bigg(3\sqrt{45}-6\sqrt{20}+10\sqrt{\dfrac{1}{5}}\bigg):\sqrt{5}\\ =\bigg(\sqrt{405}-\sqrt{720}+\sqrt{20}\bigg):\sqrt{5}\\ =\sqrt{\dfrac{405}{5}}-\sqrt{\dfrac{720}{5}}+\sqrt{\dfrac{20}{5}}\\ =\sqrt{81}-\sqrt{144}+\sqrt{4}\\ =9-12+2\\ =-1\\ 2/\\ a,\ \dfrac{1}{3+2\sqrt{2}}-\dfrac{1}{3-2\sqrt{2}}\\ =\dfrac{3-2\sqrt{2}}{(3+2\sqrt{2})(3-2\sqrt{2})}-\dfrac{3+2\sqrt{2}}{(3+2\sqrt{2})(3-2\sqrt{2})}\\ =\dfrac{3-2\sqrt{2}}{9-8}-\dfrac{3+2\sqrt{2}}{9-8}\\ =3-2\sqrt{2}-3-2\sqrt{2}\\ =(3-3)+(-2\sqrt{2}-2\sqrt{2})\\ =-4\sqrt{2}\\ b,\ \dfrac{1}{1-\sqrt{2}+\sqrt{3}}-\dfrac{1}{1-\sqrt{2}-\sqrt{3}}\\ =\dfrac{1-\sqrt{2}-\sqrt{3}}{\left [(1-\sqrt{2})+\sqrt{3}\right ].\left [(1-\sqrt{2})-\sqrt{3}\right ]}-\dfrac{1-\sqrt{2}+\sqrt{3}}{\left [(1-\sqrt{2})+\sqrt{3}\right ].\left [(1-\sqrt{2})-\sqrt{3}\right ]}\\ =\dfrac{1-\sqrt{2}-\sqrt{3}-1+\sqrt{2}-\sqrt{3}}{\left [(1-\sqrt{2})+\sqrt{3}\right ].\left [(1-\sqrt{2})-\sqrt{3}\right ]}\\ =\dfrac{-2\sqrt{3}}{(1-\sqrt{2})^2-(\sqrt{3})^2}\\ =\dfrac{-2\sqrt{3}}{3-2\sqrt{2}-3}\\ =\dfrac{2\sqrt{3}}{2\sqrt{2}}=\dfrac{\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{6}}{2}\)
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