第16页 - 通城学典课时作业本八年级数学人教版南通专版

$\frac{2}{3}$

$解：由题意得\begin{cases}1-4x≥0\\4x-1≥0\end{cases} 解得\begin{cases}x=\displaystyle{}\frac 14\\y=\displaystyle{}\frac 12\end{cases}$

$∴\frac xy=\frac 12，\frac yx=2$

$∴原式=\sqrt{\frac 12+2+2}-\sqrt{\frac 12-2+2}=\sqrt{\frac 92}-\sqrt{\frac 12}=\sqrt{2}$

$解：原式=5\sqrt{\frac 25×45}$

$=5\sqrt{2×9}$

$=15\sqrt{2}$

$解：原式=-\frac 8{10}\sqrt{6÷\frac {25}6}$

$=-\frac 45×\sqrt{ \frac {36}{25}}$

$=-\frac {24}{25}$

$解：原式=3\sqrt{3}+3×\frac {2\sqrt{3}}3-\frac {4\sqrt{3}}3-\frac 23×4\sqrt{3}$

$=3\sqrt{3}+2\sqrt{3}-\frac 43\sqrt{3}-\frac 83\sqrt{3}$

$=\sqrt{3}$

$解：原式=(\sqrt{5}+\sqrt{3})×(\sqrt{5}-\sqrt{3})×(\sqrt{5}-\sqrt{3})$

$=2\sqrt{5}-2\sqrt{3}$

$解：原式=6\sqrt{3}-4+(\sqrt{3})^2-5\sqrt{3}+3\sqrt{3}-15$

$=6\sqrt{3}-4+3-5\sqrt{3}+3\sqrt{3}-15$

$=4\sqrt{3}-16$

$解：原式=(9\sqrt 2+\frac 15×5\sqrt 2-4×\frac {\sqrt 2}2)÷4\sqrt 2$

$=8\sqrt 2÷4\sqrt 2$

$=2$

$解：原式=(2\sqrt{5})^2-(\sqrt{7})^2-(3\sqrt{2}-24)÷\sqrt{6}$

$=20-7-3×\frac {\sqrt{3}}3+4\sqrt{6}$

$=13-\sqrt{3}+4\sqrt{6}$