電子課本網 › 第121頁

第121頁

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$解：原式?=\sqrt {6}×\sqrt {2}-\sqrt {\frac {3}{8}}×\sqrt {2}?$

$? =2\sqrt {3}-\frac {\sqrt {3}}{2}?$

$? =\frac {3\sqrt {3}}{2}?$

$解：原式?=2-2\sqrt {2}+\sqrt {2}-2?$

$? =-\sqrt {2}?$

$解：原式?=(\sqrt {2})2-12?$

$? =2-1?$

$? =1?$

$解：原式?=(\sqrt {2})2-2\sqrt {6}+(\sqrt {3})2?$

$? =2-2\sqrt {6}+3?$

$? =5-2\sqrt {6}?$

$解：原式?=4a\sqrt {a}+3a\sqrt {a}-\frac {1}{2}a2×\frac {2\sqrt {a}}{a}?$

$? =7a\sqrt {a}-a\sqrt {a}?$

$? =6a\sqrt {a}?$

$解：原式?=6-4\sqrt {5}+3\sqrt {5}-10?$

$? =-4-\sqrt {5}?$

$解：原式?=(3\sqrt {2})2-(\sqrt {7})2?$

$? =18-7?$

$? =11?$

$解：原式?=(\sqrt {5})2-2×\sqrt {5}×\sqrt {10}+(\sqrt {10})2?$

$? =5-10\sqrt {2}+10?$

$? =15-10\sqrt {2}?$

$解：原式?=\frac {(\sqrt {3}-1)2}{4}?$

$? =\frac {1-2\sqrt {3}+3}{4}?$

$? =\frac {2-\sqrt {3}}{2}?$

$解：原式?=\frac {(1+\sqrt {5})2}{4}?$

$? =\frac {1+2\sqrt {5}+5}{4}?$

$? =\frac {3+\sqrt {5}}{2}?$

$解：?\frac {1}{2}×(4+\sqrt {3})×(4-\sqrt {3})=\frac {13}{2}(\mathrm {cm}2)?$

$解：?x+y=2\sqrt {3}，??x-y=2?$

$?(1)?原式?=(x+y)2=(2\sqrt {3})2=12?$

$?(2)?原式?=(x+y)(x-y)=2×2\sqrt {3}=4\sqrt {3}?$

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