第135页 - 经纶学典学霸八年级数学上下册【江苏国标】

$\frac {3\sqrt {10}}{5}$

$解:原式=-\frac{1}{3}×\sqrt{9y}=- \sqrt{y}$

$解:原式=\sqrt{\frac{2}{45}}÷ \sqrt{\frac{18}{5}}$

$=\sqrt{\frac{2}{45}×\frac{5}{18}}$

$=\frac{1}{9}$

$解:原式=\sqrt{\frac{a^{3}b^{2}}{4}}÷ \sqrt{\frac{16a^{3}}{b}}$

$=\frac{b}{8}\sqrt {b}$

$解:∵x＜1,∴y=\frac {|x-1|}{x-1}+3=\frac{-(x-1)}{x-1}+3=2$

$原式=y×\sqrt {3y}×\sqrt{y^{4}}×\sqrt{\frac{1}{y}}=\sqrt{3}y^{3}$

$当y=2时，原式=8\sqrt {3}$

$解:由题意，得\begin{cases}{ 2x-5≥0 }\ \\ { 10-4x≥0 } \end{cases}$

$解得x=\frac{5}{2},∴y=1$

$∵x\sqrt {2x}÷ \sqrt{\frac{x}{y}}=x\sqrt {2y}$

$∴当x=\frac{5}{2},y=1时，原式=\frac{5\sqrt {2}}{2}$

$解:由题意，得9-x≥0, x-6\gt 0,解得6＜x≤9$

$∵x为偶数，∴x=8$

$∴原式=\sqrt {\frac {(x+1)^{2}(x-1)^{2}}{(x+1)(x-1)}}$

$=\sqrt {9×7}=3\sqrt {7}$

$解:(1)y=2x+6可变形为2x-y+6=0$

$即A=2,B=-1,C=6$

$∴点Q(-1,3)到直线2x-y+6=0的距离$

$d=\frac{|2×(-1)-3+6|}{\sqrt{2^{2}+(-1)^{2}}}=\frac{\sqrt{5}}{5}$

$∴点Q(-1,3)直线y=2x+6的距离为\frac{\sqrt {5}}{5}$

$(2))直线x+\sqrt{6}y=5可变形为y=-\frac {\sqrt {6}}{6}x+\frac {5\sqrt {6}}{6}$

$沿y轴向上平移2个单位长度得到另一条直线y=-\frac {\sqrt {6}}{6}x+\frac {5\sqrt {6}}{6}+2$

$当x=0时,y=\frac {5\sqrt {6}}{6}+2,即点(0,\frac{5\sqrt {6}}{6}+2)$

$在直线y=-\frac{\sqrt {6}}{6}x+\frac{5\sqrt {6}}{6}+2上$

$x+ \sqrt{6}=5可变形为x+\sqrt{6}y-5=0$

$∴点(0,\frac{5\sqrt {6}}{6}+2)到直线x+\sqrt {6}y-5=0的距离$

$d=\frac {|1×0+\sqrt {6}×(\frac {5\sqrt {6}}{6}+2)-5|}{\sqrt {1^{2}+(\sqrt {6})^{2}}}=\frac{2\sqrt{42}}{7}$

$∵两直线平行，∴这两条直线之间的距离为\frac{2\sqrt{42}}{7}$