tính giúp mk vs $\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}$

0 bình luận về “tính giúp mk vs $\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}$”

1. Đáp án :

   `(\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16})/(\sqrt{2}+\sqrt{3}+\sqrt{4})=\sqrt{2}+1`

   Giải thích các bước giải :

   `(\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16})/(\sqrt{2}+\sqrt{3}+\sqrt{4})`
   `=(\sqrt{2}+\sqrt{3}+\sqrt{2.3}+\sqrt{2.4}+\sqrt{2.2.4})/(\sqrt{2}+\sqrt{3}+\sqrt{4})`
   `=(\sqrt{2}+\sqrt{3}+\sqrt{2.3}+\sqrt{2.4}+2\sqrt{2.2})/(\sqrt{2}+\sqrt{3}+\sqrt{4})`
   `=(\sqrt{2}+\sqrt{3}+\sqrt{2.2}+\sqrt{2.3}+\sqrt{2.4}+\sqrt{2.2})/(\sqrt{2}+\sqrt{3}+\sqrt{4})`
   `=[(\sqrt{2}+\sqrt{3}+\sqrt{4})+(\sqrt{2.2}+\sqrt{2.3}+\sqrt{2.4})]/(\sqrt{2}+\sqrt{3}+\sqrt{4})`
   `=[(\sqrt{2}+\sqrt{3}+\sqrt{4})+\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{4})]/(\sqrt{2}+\sqrt{3}+\sqrt{4})`
   `=[(\sqrt{2}+1)(\sqrt{2}+\sqrt{3}+\sqrt{4})]/(\sqrt{2}+\sqrt{3}+\sqrt{4})`
   `=\sqrt{2}+1`
   Vậy : `(\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16})/(\sqrt{2}+\sqrt{3}+\sqrt{4})=\sqrt{2}+1`

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2. Đáp án: `1+sqrt2`

   Giải thích các bước giải:

   `\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}`

   `=\frac{\sqrt{2}+\sqrt{3}+\sqrt{2.3}+\sqrt{2.4}+\sqrt{4}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}`

   `={(sqrt2+sqrt3+sqrt4)+sqrt2(sqrt3+sqrt4+sqrt2)}/{\sqrt{2}+\sqrt{3}+\sqrt{4}}`

   `=1+sqrt2`

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