Đáp án+Giải thích các bước giải:
\(a.\\\sqrt{75}-\sqrt{5\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2\dfrac{2}{3}}+2\sqrt{27}\\=5\sqrt{3}-\sqrt{\dfrac{16}{3}}+\dfrac{9}{2}\sqrt{\dfrac{8}{3}}+6\sqrt{3}\\= 5\sqrt{3}-\dfrac{4}{\sqrt{3}}+\dfrac{9}{2}\times \dfrac{\sqrt{8}}{\sqrt{3}}+6\sqrt{3}\\=5\sqrt{3}-\dfrac{4\sqrt{3}}{3}+\dfrac{9}{2}\times \dfrac{2\sqrt{2}}{\sqrt{3}} +6\sqrt{3}\\= 5\sqrt{3}-\dfrac{4\sqrt{3}}{3}+\dfrac{9\sqrt{2}}{\sqrt{3}}+6\sqrt{3}\\=5\sqrt{3}-\dfrac{4\sqrt{3}}{3}+3\sqrt{6}+6\sqrt{3}\\= \dfrac{15\sqrt{3}-4\sqrt{3}+18\sqrt{3}}{3}+3\sqrt{6}\\= \dfrac{29\sqrt{3}}{3}+3\sqrt{6}\\b.\\\sqrt{48}+\sqrt{5\dfrac{1}{3}}+2\sqrt{75}-5\sqrt{1\dfrac{1}{3}}\\= 4\sqrt{3}+\sqrt{\dfrac{16}{3}} +10\sqrt{3}-5\sqrt{\dfrac{4}{3}}\\= 4\sqrt{3}+\dfrac{4}{\sqrt{3}}+10\sqrt{3}-5\times \dfrac{2}{\sqrt{3}}\\= 4\sqrt{3}+\dfrac{4\sqrt{3}}{3}+10\sqrt{3}-\dfrac{10}{\sqrt{3}}\\= 4\sqrt{3}+\dfrac{4\sqrt{3}}{3}+10\sqrt{3}-\dfrac{10\sqrt{3}}{3}\\=\dfrac{12\sqrt{3}+4\sqrt{3}+30\sqrt{3}-10\sqrt{3}}{3}\\= \dfrac{36\sqrt{3}}{3}\\= 12\sqrt{3}\\c.\\(\sqrt{12}+2\sqrt{27})\dfrac{\sqrt{3}}{2}-\sqrt{150}\\=(2\sqrt{3}+6\sqrt{3})\dfrac{\sqrt{3}}{2}-\sqrt{150}\\= 8\sqrt{3}\times\dfrac{\sqrt{3}}{2}-5\sqrt{6}\\=4\sqrt{3}\times \sqrt{3}-5\sqrt{6}\\=4\times 3-5\sqrt{6}\\=12-5\sqrt{6}\\d.\\(\sqrt{18}+\sqrt{0.5} -3\sqrt{\dfrac{1}{3}})-(\sqrt{\dfrac{1}{8}}-\sqrt{75})\\= (3\sqrt{2}+\sqrt{\dfrac{1}{2}}-3\sqrt{\dfrac{1}{3}})-(\dfrac{1}{\sqrt{8}}-5\sqrt{3})\\=3\sqrt{2}+\dfrac{1}{\sqrt{2}}-3\times \dfrac{1}{\sqrt{3}}-(\dfrac{\sqrt{2}}{4}-5\sqrt{3})\\= 3\sqrt{2}+\dfrac{\sqrt{2}}{2} -\sqrt{3}-\dfrac{\sqrt{2}}{4}+5\sqrt{3}\\= \dfrac{12\sqrt{2}+2\sqrt{2}-\sqrt{2}}{4}+4\sqrt{3}\\= \dfrac{13\sqrt{2}}{4} +4\sqrt{3}\\e.\dfrac{1}{7+4\sqrt{3}}+\dfrac{1}{7-4\sqrt{3}}\\ =\dfrac{1({7-4\sqrt{3}})}{(7+4\sqrt{3})({7-4\sqrt{3}})}+\dfrac{1({7+4\sqrt{3}})}{(7-4\sqrt{3})({7+4\sqrt{3}})}\\= 7-4\sqrt{3}+7+4\sqrt{3}\\=14\\g.\\(\dfrac{1}{\sqrt{5}-\sqrt{2}}\times \dfrac{1}{\sqrt{5}+\sqrt{2}}+1)\times \dfrac{1}{(\sqrt{2}+1)^2}\\= [\dfrac{1}{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2}}+1]\times \dfrac{1}{2+2\sqrt{2}+1}\\= \dfrac{1+(\sqrt{5})-\sqrt{2})(\sqrt{5}+\sqrt{2})}{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})}\times \dfrac{1}{3+2\sqrt{2}}\\= \dfrac{4}{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})}\times \dfrac{1}{3+2\sqrt{2}}\\= \dfrac{4}{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})(3+2\sqrt{2})}\\= \dfrac{4}{(5-2)(3+2\sqrt{2})}\\=\dfrac{4(3-2\sqrt{2})}{3}\\= \dfrac{12-8\sqrt{2}}{3}\)
