9.(2025·甘肃改编)计算$\sqrt{12}÷(\sqrt{6}×\frac{1}{\sqrt{2}})$的结果为
2
.
答案：9. 2
10.化简:$-\frac{15}{8}\sqrt{2\frac{10}{27}}÷\sqrt{\frac{25}{12a^{3}}}=$
$-2a\sqrt{a}$
.
答案：10. $-2a\sqrt{a}$
11.计算:(1)$3\sqrt{1\frac{1}{7}}÷\frac{\sqrt{2}}{2}×\sqrt{\frac{3}{14}}$；
(2)$3\sqrt{1.25}÷\frac{3}{4}\sqrt{2\frac{1}{2}}×2\sqrt{18}$；
(3)$\frac{3}{5}\sqrt{ab^{2}}·(-\frac{5}{6}\sqrt{a^{3}b})÷\frac{4}{15}\sqrt{\frac{b^{2}}{a}}$；
(4)$6a\sqrt{a^{2}b^{5}}÷(-2\sqrt{a^{3}b})×\sqrt{\frac{b}{a}}$.
答案：11.解:(1)原式$=3×\sqrt{\frac{8}{7}}×\frac{2}{\sqrt{2}}×\frac{\sqrt{3}}{\sqrt{7}×\sqrt{2}}$
$=3×\frac{2\sqrt{2}}{\sqrt{7}}×\frac{2}{\sqrt{2}}×\frac{\sqrt{3}}{\sqrt{7}×\sqrt{2}}=\frac{6}{7}\sqrt{6}$.
(2)原式$=3\sqrt{\frac{5}{4}}÷\frac{3}{4}\sqrt{\frac{5}{2}}×2\sqrt{18}$
$=3\sqrt{\frac{5}{4}}×\frac{4}{3}\sqrt{\frac{2}{5}}×2\sqrt{18}$
$=3×\frac{4}{3}×2×\sqrt{\frac{5}{4}×\frac{2}{5}×18}=24$.
(3)原式$=\frac{3}{5}b\sqrt{a}·(-\frac{5}{6}a\sqrt{ab})÷\frac{4b}{15\sqrt{a}}$
$=\frac{3}{5}b\sqrt{a}·(-\frac{5}{6}a\sqrt{ab})·\frac{15\sqrt{a}}{4b}$
$=-(\frac{3}{5}b·\frac{5}{6}a·\frac{15}{4b})·(\sqrt{a}·\sqrt{ab}·\sqrt{a})$
$=-\frac{15}{8}a· a\sqrt{ab}=-\frac{15}{8}a^{2}\sqrt{ab}$.
(4)原式$=-3a\sqrt{\frac{a^{2}b^{5}}{a^{3}b}×\frac{b}{a}}$
$=-3a\sqrt{\frac{b^{5}}{a^{2}}}=-3\sqrt{\frac{b^{5}}{a^{2}}× a^{2}}=-3\sqrt{b^{5}}=-3b^{2}\sqrt{b}$.
12.已知$\sqrt{\frac{x-6}{9-x}}=\frac{\sqrt{x-6}}{\sqrt{9-x}}$,且$x$为奇数,求$(x+1)$的值.
答案：12.解:由分式和二次根式有意义的条件,得$\begin{cases} x-6≥0, \\9-x＞0, \end{cases}$
解得$6≤ x＜9$,$\because x$为奇数,$\therefore x=7$,
$\therefore$原式$=(x+1)\sqrt{\frac{(x-1)^{2}}{(x+1)(x-1)}}=(x+1)\sqrt{\frac{x-1}{x+1}}$
$=\sqrt{(x+1)(x-1)}=\sqrt{(7+1)×(7-1)}=4\sqrt{3}$.
13.(2025·海淀区期中)定义:若两个二次根式$m,n$满足$m· n=p$,且$p$是有理数,则称$m$与$n$是关于$p$的和谐二次根式.
(1)若$m$与$\sqrt{3}$是关于6的和谐二次根式,求$m$的值；
(2)若$2-\sqrt{2}$与$4+\sqrt{2}m$是关于4的和谐二次根式,求$m$的值.
答案：13.解:(1)由题意得$\sqrt{3}m=6$,$\therefore m=\frac{6}{\sqrt{3}}=\frac{2×3}{\sqrt{3}}=2\sqrt{3}$.
(2)由题意得$(2-\sqrt{2})(4+\sqrt{2}m)=4$,
$(\sqrt{2})^{2}·(\sqrt{2}-1)(2\sqrt{2}+m)=4$,
整理,得$(\sqrt{2}-1)m=2\sqrt{2}-2$,$\therefore m=2$.
