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$\sqrt{6}$
解：原式​$=4\sqrt {3}+\frac {\sqrt {2}}{2}+\frac {\sqrt {3}}{4}-5\sqrt {2}$​
​$=\frac {17\sqrt {3}}{4}-\frac {9\sqrt {2}}{2}$​
$ \begin{aligned}解:原式&=2\sqrt 6-\frac {\sqrt 3}{3}-\frac {\sqrt 3}{9}-\sqrt 6 \\ &=\sqrt{6}-\frac{4\sqrt{3}}{9} \\ \end{aligned}$
$ \begin{aligned} 解:原式&=2\sqrt x-3\sqrt x+2\sqrt x \\ &=\sqrt x \\ \end{aligned}$
$ \begin{aligned} 解:原式&=\frac 32\sqrt {2x}+5\sqrt {2x}-\frac 12\sqrt {2x} \\ &=6 \sqrt{2x} \\ \end{aligned}$
解：$\because a=\frac{1}{2 + \sqrt{3}}=\frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})}=2 - \sqrt{3}＜1，$
$\therefore$原式$=\frac{(a - 1)^{2}}{a - 1}+\frac{\sqrt{(a - 1)^{2}}}{a(a - 1)}$
$=a - 1+\frac{1 - a}{a(a - 1)}=a - 1-\frac{1}{a}$
$=2 - \sqrt{3}-1-\frac{1}{\frac{1}{2 + \sqrt{3}}}=2 - \sqrt{3}-1-(2 + \sqrt{3})$
$=-1 - 2\sqrt{3}。$
解：由题意得$\begin{cases}3a + 4 = 19 - 2a\\4a - 3x = 0\\y - a = 0\end{cases}，$
解$3a + 4 = 19 - 2a，$
移项可得$3a+2a=19 - 4，$
$5a = 15，$解得$a = 3。$
把$a = 3$代入$4a - 3x = 0，$得$4\times3 - 3x = 0，$
$12-3x = 0，$$3x = 12，$解得$x = 4。$
把$a = 3$代入$y - a = 0，$得$y - 3 = 0，$解得$y = 3。$
$\therefore$原式$=(4\times1\div10)\cdot\sqrt{xy^{2}}\cdot\frac{x}{y^{3}}\div\frac{y}{x^{2}}$
$=\frac{2}{5}\sqrt{xy^{2}}\cdot\frac{x}{y^{3}}\cdot\frac{x^{2}}{y}=\frac{2}{5}\sqrt{\frac{x^{4}}{y^{2}}}=\frac{2x^{2}}{5y}=\frac{2\times4^{2}}{5\times3}=\frac{32}{15}。$
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