第105页 - 创新优化学案八年级数学苏科版

解：原式$=4\sqrt 2-2\sqrt 3+2\sqrt 3-(3\sqrt 2-3\sqrt 3)$

$ =(4-3)×\sqrt 2+(-2+2+3)×\sqrt 3$

$ =\sqrt 2+3\sqrt 3$

解：原式$=6\sqrt {10}-\frac {\sqrt {10}}5-\frac {\sqrt {10}}5$

$ =\frac {28\sqrt {10}}5$

解：原式$=4\sqrt 5+3\sqrt 5-2\sqrt 2+4\sqrt 2$

$ =7\sqrt 5+2\sqrt 2$

解：原式$=3+1+2\sqrt 3-2\sqrt 3+\frac {2\sqrt 3}3$

$ =4+\frac {2\sqrt 3}3$

解：原式$=\sqrt 6-\sqrt 6+\sqrt 6-2\sqrt 6$

$ =-\sqrt 6$

解：原式$=14a\sqrt {2a}+7a\sqrt {2a}-4a^2\frac {\sqrt {8a}}{8a}$

$ =21a\sqrt {2a}-a\sqrt {2a}$

$ =20a\sqrt {2a}$

解$ ∶ (1) $∵$5 \sqrt {\frac x 5}+\frac 12 \sqrt {20x}+\frac 54 x \sqrt {\frac 4{5x}}$

$ =5 ×\frac {\sqrt {5x}}5+\frac 12 ×2 \sqrt {5x}+\frac 54x ×\frac {2 \sqrt {5x}}{5x}$

$ =\sqrt {5x}+\sqrt {5x}+\frac 12 \sqrt {5x}$

$ =\frac 52 \sqrt {5x}$

∴这个三角形的周长是$ \frac 52 \sqrt {5x} $

$ (2)$当$ x=20 $时，$\frac 52 \sqrt {5x}=\frac 52 \sqrt {5×20}=\frac 52 ×10=25$

解：$(1) \sqrt {13-2 \sqrt {42}} $

$ =\sqrt {7-2 ×\sqrt {7 ×6}+6} $

$ =\sqrt {(\sqrt 7)^2-2 ×\sqrt 7 ×\sqrt 6+(\sqrt 6)^2} $

$ =\sqrt {(\sqrt 7-\sqrt 6)^2} $

$ =\sqrt 7-\sqrt 6； $

$ (2)\sqrt {7-\sqrt {40}} $

$ =\sqrt {7-2 \sqrt {10}} $

$ =\sqrt {5-2 \sqrt {5 ×2}+2} $

$ =\sqrt {(\sqrt 5)^2-2 ×\sqrt 5 ×\sqrt 2+(\sqrt 2)^2} $

$ =\sqrt {(\sqrt 5-\sqrt 2)^2} $

$ =\sqrt 5-\sqrt 2； $

$ (3)\sqrt {2-\sqrt 3} $

$ =\sqrt {\frac {8-4 \sqrt 3}4} $

$ =\sqrt {\frac {8-2 \sqrt {12}}4} $

$ =\sqrt {\frac {6-2 \sqrt {6 ×2}+2}4} $

$ =\sqrt {\frac {(\sqrt 6)^2-2 ×\sqrt 6 ×\sqrt 2+(\sqrt 2)^2}4} $

$ =\sqrt {\frac {(\sqrt 6-\sqrt 2)^2}4} $

$ =\frac {\sqrt 6-\sqrt 2}2 . $