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電子課本網(wǎng) › 第105頁

第105頁

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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}$?
∴這個三角形的周長是?$ \frac 52 \sqrt {5x} $?
?$ (2)$?當(dāng)?$ x=20 $?時，?$\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 . $?
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