一、填空题
1. $\sqrt{24}-\sqrt{\dfrac{6}{5}}×\sqrt{45} =$
2. $\sqrt{2}×(\sqrt{18}-\dfrac{1}{2}\sqrt{8}) =$

1. $-\sqrt{6}$ 2. $4$

3. $(\sqrt{27}+\sqrt{18})(\sqrt{3}-\sqrt{2})=$
4. $(2\sqrt{3}-\sqrt{2})^2=$

3. $3$ 4. $14-4\sqrt{6}$

5. 一题多解 $(\sqrt{\dfrac{4}{3}}+\sqrt{3})×\sqrt{6} =$
6. $\sqrt{24}÷\sqrt{3}-\dfrac{6}{\sqrt{2}}+\sqrt{32} =$

5. $5\sqrt{2}$
解析：解法一 原式$=\sqrt{\dfrac{4}{3}}×\sqrt{6}+\sqrt{3}×\sqrt{6} = \sqrt{\dfrac{4}{3}×6}+\sqrt{3×6}=2\sqrt{2}+3\sqrt{2}=5\sqrt{2}.$
解法二 原式$=(\dfrac{2\sqrt{3}}{3}+\sqrt{3})×\sqrt{6}=\dfrac{5\sqrt{3}}{3}×\sqrt{6}=5\sqrt{2}.$
6. $3\sqrt{2}$

二、解答题
7. $(\sqrt{3}-2)^2 + \sqrt{12} + 6\sqrt{\dfrac{1}{3}}$
8. $(\sqrt{18}-\sqrt{12}+\sqrt{2}) × 2\sqrt{6}$
9. $(\sqrt{3}+\sqrt{2}+\sqrt{5})(\sqrt{3}-\sqrt{2}-\sqrt{5})$
10. $(3+2\sqrt{2})(3-2\sqrt{2}) + (1-\sqrt{2})^2$

7. $7$
8. $16\sqrt{3}-12\sqrt{2}$
9. $-4-2\sqrt{10}$
10. $4-2\sqrt{2}$

11. 先化简,再求值:$\sqrt{25xy}+x\sqrt{\frac{y}{x}}-4y\sqrt{\frac{x}{y}}-\frac{1}{y}\sqrt{xy^3}$,其中$x=\frac{1}{3},y=4$.

11. 原式$=\sqrt{xy}$. 当 $x = \dfrac{1}{3}, y = 4$ 时, 原式$=\sqrt{\dfrac{1}{3}×4}=\dfrac{2\sqrt{3}}{3}$

12. 一题多解 已知$x=\frac{\sqrt{5}-1}{2}, y=\frac{\sqrt{5}+1}{2}$,求$x^2+xy+y^2$的值.

12. 解法一 由题意,得 $x+y=\dfrac{\sqrt{5}-1}{2}+\dfrac{\sqrt{5}+1}{2}=\sqrt{5},xy=\dfrac{\sqrt{5}-1}{2}×\dfrac{\sqrt{5}+1}{2}=1.$ 所以 $x^2+xy+y^2=x^2+2xy+y^2-xy=(x+y)^2-xy=(\sqrt{5})^2-1=4.$
解法二 原式$=(\dfrac{\sqrt{5}-1}{2})^2 +\dfrac{\sqrt{5}-1}{2}×\dfrac{\sqrt{5}+1}{2}+(\dfrac{\sqrt{5}+1}{2})^2 =\dfrac{5-2\sqrt{5}+1}{4}+\dfrac{5-1}{4}+\dfrac{5+2\sqrt{5}+1}{4}=4.$