a/ $\sqrt{12}-2\sqrt{27}+3\sqrt{75}-\sqrt{48}\\=\sqrt{4.3}-2\sqrt{9.3}+3\sqrt{25.3}-\sqrt{16.3}\\=\sqrt 4.\sqrt 3-2.\sqrt 9.\sqrt 3+3.\sqrt{25}.\sqrt 3-\sqrt{16}.\sqrt 3\\=2\sqrt 3-2.3.\sqrt 3+3.5.\sqrt 3-4\sqrt 3\\=-2\sqrt 3-6\sqrt 9+15\sqrt 3\\=7\sqrt 3$
Vậy $\sqrt{12}-2\sqrt{27}+3\sqrt{75}-\sqrt{48}=7\sqrt 3$
b/ $1+2\sqrt{27}-\sqrt{4+2\sqrt 3}\\=1+2.\sqrt{9.3}-\sqrt{3+2\sqrt 3+1}\\=1+2.\sqrt 9.\sqrt 3-\sqrt{(\sqrt 3+1)^2}\\=1+2.3.\sqrt 3-|\sqrt 3+1|\\=1+6\sqrt 3-(\sqrt 3+1)(vì\,\,\sqrt 3+1>0)\\=1+6\sqrt 3-\sqrt 3-1\\=5\sqrt 3$
Vậy $1+2\sqrt{27}-\sqrt{4+2\sqrt 3}=5\sqrt 3$
c/ $\sqrt{32}+\dfrac{\sqrt{62}}{\sqrt{31}}-12\sqrt{\dfrac 1 2}+\dfrac{6}{\sqrt 2}\\=\sqrt{16.2}+\dfrac{\sqrt{31.2}}{\sqrt{31}}-\sqrt{144}.\sqrt{\dfrac 1 2}+\dfrac{\sqrt{36}}{\sqrt 2}\\=\sqrt{16}.\sqrt 2+\dfrac{\sqrt{31}.\sqrt 2}{\sqrt{31}}-\sqrt{\dfrac{144}{2}}+\dfrac{\sqrt{2.18}}{\sqrt 2}\\=4\sqrt 2+\sqrt 2-\sqrt{72}+\dfrac{\sqrt2.\sqrt{18}}{\sqrt 2}\\=5\sqrt 2-\sqrt{36.2}+\sqrt{18}\\=5\sqrt 2-\sqrt{36}.\sqrt 2+\sqrt{9.2}\\=5\sqrt 2-6\sqrt 2+\sqrt 9.\sqrt 2\\=-\sqrt 2+3\sqrt 2\\=2\sqrt 2$
Vậy $\sqrt{32}+\dfrac{\sqrt{62}}{\sqrt{31}}-12\sqrt{\dfrac{1}{2}}+\dfrac{6}{\sqrt 2}=2\sqrt 2$
d/ $\dfrac{2}{\sqrt 3-1}+\dfrac{2}{\sqrt 3+1}+\dfrac{3-\sqrt 3}{1-\sqrt 3}\\=\dfrac{2(\sqrt 3+1)}{(\sqrt 3-1)(\sqrt 3+1)}+\dfrac{2(\sqrt 3-1)}{(\sqrt 3+1)(\sqrt 3-1)}+\dfrac{(\sqrt 3)^2-\sqrt 3}{1-\sqrt 3}\\=\dfrac{2(\sqrt 3+1)+2(\sqrt 3-1)}{(\sqrt 3+1)(\sqrt 3-1)}-\dfrac{\sqrt 3(\sqrt 3-1)}{\sqrt 3-1}\\=\dfrac{2\sqrt 3+2+2\sqrt 3-2}{(\sqrt 3)^2-1^2}-\sqrt 3\\=\dfrac{4\sqrt 3}{3-1}-\sqrt 3\\=\dfrac{4\sqrt 3}{2}-\sqrt 3\\=2\sqrt 3-\sqrt 3\\=\sqrt 3$
Vậy $\dfrac{2}{\sqrt 3-1}+\dfrac{2}{\sqrt 3+1}+\dfrac{3-\sqrt 3}{1-\sqrt 3}=\sqrt 3$
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