7. 若$a = \sqrt{2}$,$b = \sqrt{7}$,则$\sqrt{\frac{14a^{2}}{b^{2}}}$ =()

A.2
B.4
C.$\sqrt{7}$
D.$\sqrt{2}$

A

8. 若$\sqrt{3a + 4}$是最简二次根式,且$a$为整数,则$a$的最小值是.

1

9. 对于任意两个和为正数的实数$a$,$b$,定义运算※:$a※b = \frac{a - b}{\sqrt{a + b}}$,例如:$3※1 = \frac{3 - 1}{\sqrt{3 + 1}} = 1$,那么$8※12$ =.

$-\frac{2\sqrt{5}}{5}$

10. 已知$xy ＞ 0$,则化简$x\sqrt{-\frac{y}{x^{2}}}$的结果是.

$-\sqrt{-y}$

11. 计算:
(1)$\sqrt{\frac{20x^{2}y^{2}}{z^{5}}}(x≥ 0,y≥ 0)$;
(2)$\sqrt{27a^{2}b^{3}c}÷\sqrt{3a^{2}bc^{2}}·\sqrt{\frac{c^{3}}{b^{2}}}(c ＞ 0)$;
(3)$\frac{2}{b}\sqrt{ab^{5}}÷3\sqrt{\frac{b}{a}}·(-\frac{3}{2}\sqrt{a^{2}b})(a ＞ 0,b ＞ 0)$.

解：原式$​=\frac {\sqrt {20x^2y^2}}{\sqrt {z^5}}​$
$​=\frac {2xy\sqrt {5z}}{z^3}​$
解：原式$​=\sqrt {27a^2b^3c÷3a^2bc^2}·\sqrt {\frac {c^3}{b^2}}​$
$​=\sqrt {9c^2}​$
​=3c​
解：原式$​=\frac {2}{b}×\frac {1}{3}×(-\frac {3}{2})\sqrt {ab^5·\frac {a}{b}· a^2b}​$
$​=-\frac {1}{b}\sqrt {a^4b^5}​$
$​=-a^2b\sqrt {b}​$

12. 在学完"二次根式的乘除"后,数学老师给同学们留下这样一道思考题:已知$x + y = - 6$,$xy = 4$,求$\sqrt{\frac{y}{x}}+\sqrt{\frac{x}{y}}$的值.
小刚是这样解的:$\sqrt{\frac{y}{x}}+\sqrt{\frac{x}{y}} = \frac{\sqrt{y}}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{y}} = \frac{\sqrt{xy}}{x}+\frac{\sqrt{xy}}{y} = \frac{\sqrt{xy}(x + y)}{xy}$.
把$x + y = - 6$,$xy = 4$代入,得$\frac{\sqrt{xy}(x + y)}{xy} = \frac{\sqrt{4}×(-6)}{4} = - 3$.
显然,这个解法是错误的,请你写出正确的解题过程.

解：$\because x+y=-6,xy=4,\therefore x＜0,y＜0,$
$ \therefore\sqrt{\frac{y}{x}}+\sqrt{\frac{x}{y}}=-\frac{\sqrt{xy}}{x}-\frac{\sqrt{xy}}{y}=-\frac{\sqrt{xy}(x+y)}{xy}.$
把x+y=-6,xy=4代入，
得$-\frac{\sqrt{xy}(x+y)}{xy}=-\frac{\sqrt{4}×(-6)}{4}=3.$