CMR: a/ $\sqrt{10+\sqrt{60}-\sqrt{24}-\sqrt{40}=}\sqrt{3}+\sqrt{5}-\sqrt{2}$ b/ $\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}-\sqrt{3}=\sqrt{2}+1$

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1. Đáp án:

   Giải thích các bước giải:

   a/ $\sqrt{10+\sqrt{60}-\sqrt{24}-\sqrt{40}}$

   $=\sqrt{(\sqrt{2})^2+(\sqrt{3})^2+(\sqrt{5})^2+2\sqrt{3}.\sqrt{5}-2\sqrt{2}.\sqrt{3}-2\sqrt{2}.\sqrt{5}}$

   $=\sqrt{(\sqrt{3}+\sqrt{5}-\sqrt{2})^2}$

   $=\sqrt{3}+\sqrt{5}-\sqrt{2}$

   b/ $\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}-\sqrt{3}$

   $=\sqrt{(\sqrt{2})^2+(\sqrt{3})^2+1+2\sqrt{2}.\sqrt{3}+2.1\sqrt{3}+2.1\sqrt{2}}-\sqrt{3}$

   $=\sqrt{(\sqrt{2}+\sqrt{3}+1)^2}-\sqrt{3}$

   $=\sqrt{2}+\sqrt{3}+1-\sqrt{3}$

   $=\sqrt{2}+1$

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2. a) Ta có:

   $10 + \sqrt{60} -\sqrt{24} – \sqrt{40}$

   $= 5 + 3 + 2 + 2\sqrt{15} – 2\sqrt{6} – 2\sqrt{10}$

   $= (\sqrt5)^2 + (\sqrt3)^2 + (\sqrt2)^2 + 2\sqrt3.\sqrt5 – 2\sqrt2.\sqrt3 – 2\sqrt2.\sqrt5$

   $= (\sqrt2 -\sqrt5 – \sqrt3)^2$

   $= (\sqrt3 + \sqrt5 – \sqrt2)^2$

   Do đó:

   $\sqrt{10 + \sqrt{60} -\sqrt{24} – \sqrt{40}} = \sqrt3 + \sqrt5 -\sqrt2$

   b) Ta có:

   $6 + \sqrt{24} + \sqrt{12} + \sqrt8$

   $= 1 + 2 + 3 + 2\sqrt6 + 2\sqrt3 + 2\sqrt2$

   $= 1^2 + (\sqrt2)^2 + (\sqrt3)^2 + 2\sqrt2.\sqrt3 + 2\sqrt3.1 + 2\sqrt2.1$

   $= (1 + \sqrt2 + \sqrt3)^2$

   Do đó:

   $\sqrt{6 + \sqrt{24} + \sqrt{12} + \sqrt8} = 1 + \sqrt2 + \sqrt3$

   Hay $\sqrt{6 + \sqrt{24} + \sqrt{12} + \sqrt8} -\sqrt3 = \sqrt2 + 1$

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