一、填空题
1. $\dfrac{14}{\sqrt{7}} - \sqrt{28} =$
2. $(2\sqrt{12} - 3\sqrt{\dfrac{1}{3}}) × \sqrt{6} =$
3. $\dfrac{\sqrt{3}}{\sqrt{3} + \sqrt{12}} =$
4. [天津中考] $(\sqrt{61} + 1)(\sqrt{61} - 1) =$
5. $(2\sqrt{3} + \sqrt{6})(2\sqrt{3} - \sqrt{6}) =$
6. $(3\sqrt{2} - 2\sqrt{5})(2\sqrt{5} + 3\sqrt{2}) =$

1. 0
2. $9\sqrt{2}$
3. $\dfrac{1}{3}$
4. 60
5. 6
6. -2

二、计算题
7. [青岛中考] $\frac{\sqrt{18}+\sqrt{50}}{\sqrt{2}} - (\sqrt{3})^0$
8. $(\sqrt{96} - \sqrt{\frac{1}{6}}) ÷ \sqrt{3}$
9. $2(\sqrt{12} + \sqrt{20}) - 3(\sqrt{3} - \sqrt{5})$
10. $(\sqrt{3} - 2\sqrt{5})(\sqrt{15} +5) - (\sqrt{10} - \sqrt{2})^2$
11. $(4\sqrt{2} -3\sqrt{6}) ÷ \sqrt{2} +12\sqrt{\frac{1}{3}}$
12. $(2\sqrt{3} + \sqrt{6}) × (2\sqrt{3} - \sqrt{6}) - (\sqrt{3} - \sqrt{2})^2$
13. $(2\sqrt{2} -3)^{2024} × (2\sqrt{2} +3)^{2025}$
14. 一题多解 $(1+\sqrt{2})^2(1+\sqrt{3})^2(1-\sqrt{2})^2(1-\sqrt{3})^2$

7. 7
8. $\dfrac{23\sqrt{2}}{6}$
9. $\sqrt{3}+7\sqrt{5}$
10. $-3\sqrt{5}-5\sqrt{3}-12$
11. $4+\sqrt{3}$
12. $1+2\sqrt{6}$
13. $2\sqrt{2}+3$
14. 解法一 原式=$(3+2\sqrt{2})(4+2\sqrt{3})(3-2\sqrt{2})(4-2\sqrt{3}) = [(3+2\sqrt{2})(3-2\sqrt{2})][(4+2\sqrt{3})(4-2\sqrt{3})] = 1×4=4.$
解法二 原式=$[(1+\sqrt{2})^2(1-\sqrt{2})^2][(1+\sqrt{3})^2(1-\sqrt{3})^2] = [(1+\sqrt{2})(1-\sqrt{2})]^2[(1+\sqrt{3})(1-\sqrt{3})]^2 = (-1)^2×(-2)^2=4.$