Đáp án:

 `a)0`

`b)0`

`c)0`

Giải thích các bước giải:

 `a)\frac{\sqrt{10}-\sqrt{6}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{14}+\sqrt{6}}{\sqrt{7}+\sqrt{3}}`

`=\frac{\sqrt{2}.\sqrt{5}-\sqrt{2}.\sqrt{3}}{\sqrt[5}-\sqrt{3}}-\frac{\sqrt{2}.\sqrt{7}+\sqrt{2}.\sqrt{3}}{\sqrt{7}+\sqrt{3}}`

`=\frac{\sqrt{2}.(\sqrt{5}-\sqrt{3})}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{2}.(\sqrt{7}+\sqrt{3})}{\sqrt{7}+\sqrt{3}}`

`=\sqrt{2}-\sqrt{2}`

`=0`

`b)\frac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}+\frac{\sqrt{15}+\sqrt{6}}{\sqrt{5}+\sqrt{2}}-\sqrt{12}`

`=\frac{\sqrt{3}.\sqrt{2}-\sqrt{3}.1}{\sqrt{2}-1}+\frac{\sqrt{3}.\sqrt{5}+\sqrt{3}.\sqrt{2}}{\sqrt{5}+\sqrt{2}}-\sqrt{2^{2}.3}`

`=\frac{\sqrt{3}.(\sqrt{2}-1)}{\sqrt{2}-1}+\frac{\sqrt{3}.(\sqrt{5}+\sqrt{2})}{\sqrt{5}+\sqrt{2}}-2\sqrt{3}`

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

`=0`

`c)\frac{\sqrt{21}-\sqrt{14}}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{28}-\sqrt{35}}{2-\sqrt{5}}-\sqrt{28}`

`=\frac{\sqrt{7}.\sqrt{3}-\sqrt{7}.\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{7}.\sqrt{4}-\sqrt{7}.\sqrt{5}}{2-\sqrt{5}}-\sqrt{2^{2}.7}`

`=\frac{\sqrt{7}.(\sqrt{3}-\sqrt{2})}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{7}.(2-\sqrt{5})}{2-\sqrt{5}}-2\sqrt{7}`

`=\sqrt{7}+\sqrt{7}-2\sqrt{7}`

`=0`