Rút gọn: 1/ 2 + √3 + √2 / √6 - 2 / 3 + √3

2 câu trả lời

`#tnvt`

`\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}`

`=\frac{2-\sqrt{3}}{4-3}+\frac{\sqrt{2}}{\sqrt{3}.\sqrt{2}}-\frac{2(3-\sqrt{3})}{9-3}`

`=2-\sqrt{3}+\frac{1}{\sqrt{3}}-\frac{3-\sqrt{3}}{3}`

`=2-\sqrt{3}+\frac{1}{\sqrt{3}}-\frac{\sqrt{3}(\sqrt{3}-1)}{3}`

`=2-\sqrt{3}+\frac{1-\sqrt{3}+1}{\sqrt{3}}`

`=\frac{(2-\sqrt{3}).\sqrt{3}+2-\sqrt{3}}{\sqrt{3}}`

`=\frac{2\sqrt{3}-3+2-\sqrt{3}}{\sqrt{3}}`

`=\frac{\sqrt{3}-1}{\sqrt{3}}`

`=\frac{3-\sqrt{3}}{3}`

 

Đáp án: `(3-sqrt3)/3` 

Giải thích các bước giải:

\(\dfrac{1}{2+\sqrt3}+\dfrac{\sqrt2}{\sqrt6} - \dfrac{2}{3+\sqrt3}\\=\dfrac{2-\sqrt3}{(2+\sqrt3)(2-\sqrt3)}+\dfrac{\sqrt2}{\sqrt2 . \sqrt3}-\dfrac{2(3-\sqrt3)}{(3+\sqrt3)(3-\sqrt3)}\\=\dfrac{2-\sqrt3}{1}+\dfrac{1}{\sqrt3}-\dfrac{2(3-\sqrt3)}{6}\\=2-\sqrt3+\dfrac{\sqrt3}{3}-\dfrac{3-\sqrt3}{3}=\dfrac{3(2-\sqrt3)+\sqrt3-(3-\sqrt3)}{3}=\dfrac{6-3\sqrt3+\sqrt3-3+\sqrt3}{3}=\dfrac{3-\sqrt3}{3}=\dfrac{\sqrt3.(\sqrt3-1)}{3}\\=\dfrac{\sqrt3-1}{\sqrt3}=\dfrac{\sqrt3(\sqrt3-1)}{3}=\dfrac{3-\sqrt3}{3}\)

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