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電子課本網(wǎng) › 第124頁

第124頁

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$解：原式?=\sqrt {40×10}?$

$? =\sqrt {400}?$

$? =20?$

$解：原式?=\sqrt {50÷2}?$

$? =\sqrt {25}?$

$? =5?$

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

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

$解：原式?=3-2\sqrt {11}+3\sqrt {11}-22?$

$? =\sqrt {11}-19?$

$解：原式?=(\sqrt {2})2-(\sqrt {5})2?$

$? =2-5?$

$? =-3?$

$解：原式?=3-2\sqrt {6}+2?$

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

$解：原式?=8\sqrt {3}×\frac {\sqrt {2}}{2}÷5\sqrt {3}?$

$? =4\sqrt {6}×\frac {\sqrt {3}}{15}?$

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

$解：原式?=14a\sqrt {2a}-4a2×\frac {1}{2\sqrt {2a}}+7a\sqrt {2a}?$

$? =14a\sqrt {2a}-a\sqrt {2a}+7a\sqrt {2a}?$

$? =20a\sqrt {2a}?$

$解：將?a=2，??b=-8，??c=5?代入得：$

$原式?=\frac {-8+\sqrt {64-4×2×5}}{2×2}=\frac {8+2\sqrt {6}}{4} =2+\frac {\sqrt {6}}{2}?$

$解：原式?=a2+2ab+b2+(2a2+ab-2ab-b2)-3a2=ab?$

$將?a=2+\sqrt {3}，??b=2-\sqrt {3}?代入原式得：$

$原式?=(2+\sqrt {3})(2-\sqrt {3}) =4-3 =1?$

$解：菱形的邊長為?\sqrt {96}÷4=\sqrt {6}(\mathrm {cm})?$

$∵?∠DAB = 120°?$

$∴?∠DAO= 60°?$

$∴?OA=\frac {\sqrt {6}}{2}\mathrm {cm}，??OD=\frac {3\sqrt {2}}{2}\mathrm {cm}?$

$∴?AC=\sqrt {6}(\mathrm {cm})，??BD = 3\sqrt {2}(\mathrm {cm})?$

$∴?S=\frac {1}{2}×AC×BD=3\sqrt {3}(\mathrm {cm}2)?$

$解：∵?(a+\frac {1}{a})2=a2+2+\frac {1}{a2}=10?$

$∴?a2+\frac {1}{a2}=8?$

$?(a-\frac {1}{a})2=a2-2+\frac {1}{a2}=6?$

$∴?a-\frac {1}{a}=±\sqrt {6}?$