Đáp án:
Giải thích các bước giải:
`\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}`
`=\frac{2\sqrt{3}.\sqrt{5}-2\sqrt{2}.\sqrt{5}+\sqrt{2}.\sqrt{3}-\sqrt{3}.\sqrt{3}}{2\sqrt{5}-2\sqrt{2}.\sqrt{5}-\sqrt{3}+\sqrt{2}.\sqrt{3}}`
`=\frac{2\sqrt{5}(\sqrt{3}-\sqrt{2})+\sqrt{3}(\sqrt{2}-\sqrt{3})}{2\sqrt{5}(1-\sqrt{2})-\sqrt{3}(1-\sqrt{2})}`
`=\frac{2\sqrt{5}(\sqrt{3}-\sqrt{2})-\sqrt{3}(\sqrt{3}-\sqrt{2})}{2\sqrt{5}(1-\sqrt{2})-\sqrt{3}(1-\sqrt{2})}`
`=\frac{(2\sqrt{5}-3)(\sqrt{3}-\sqrt{2})}{(2\sqrt{5}-3)(1-\sqrt{2})}`
`=\frac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}`
