Tính $\frac{6}{\sqrt{2}-\sqrt{3}+3}$

Đáp án+Giải thích các bước giải:

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

`=\frac{6(\sqrt{2}-\sqrt{3}-3)}{(\sqrt{2}-\sqrt{3})^2-3^2}`

`=\frac{6\sqrt{2}-6\sqrt{3}-18}{3-2\sqrt{6}+2-9}`

`=\frac{6\sqrt{2}-6\sqrt{3}-18}{-2\sqrt{6}-4}`

`=\frac{-2(3\sqrt{2}+3\sqrt{3}+9)}{-2(\sqrt{6}+2)}`

`=\frac{-3\sqrt{2}+3\sqrt{3}+9}{\sqrt{6}+2}`

`=\frac{(-3\sqrt{2}+3\sqrt{3}+9)(\sqrt{6}-2)}{6-4}`

`=\frac{-6\sqrt{3}+6\sqrt{2}+9\sqrt{2}-6\sqrt{3}+9\sqrt{6}-18}{2}`

`=\frac{-12\sqrt{3}+15\sqrt{2}+9\sqrt{6}-18}{2}`

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

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

`=\frac{6(3+\sqrt{2}+\sqrt{3})}{(3+\sqrt{2})^2-\sqrt{3}^2}`

`=\frac{18+6\sqrt{2}+6\sqrt{3}}{9+6\sqrt{2}+2-3}`

`=\frac{18+6\sqrt{2}+6\sqrt{3}}{8+6\sqrt{2}}`

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

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

`=\frac{36-27\sqrt{2}+12\sqrt{2}-18+12\sqrt{3}-9\sqrt{6}}{16-18}`

`=\frac{-18+15\sqrt{2}+9\sqrt{6}-12\sqrt{3}}{2}`