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第119頁(yè)

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A

C

$解：原式?=3\sqrt {3}?$

$解：原式?=3\sqrt {2}-\sqrt {2}?$

$? =2\sqrt {2}?$

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

$? =\frac {1}{3}\sqrt {ab}?$

$解：原式?=10\sqrt {3}-4\sqrt {2}-9\sqrt {3}?$

$? =\sqrt {3}-4\sqrt {2}?$

$解：原式?=6\sqrt {2}-5\sqrt {2}+\frac {1}{3}×3\sqrt {5}?$

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

$解：原式?=6\sqrt {5}+9\sqrt {5}-6\sqrt {5}?$

$? =9\sqrt {5}?$

$解：原式?=\sqrt {3a}+3\sqrt {a}-\sqrt {3a}?$

$? =3\sqrt {a}?$

$解：原式?=2\sqrt {3}-10\sqrt {2}-5\sqrt {3}+3\sqrt {2}?$

$? =-3\sqrt {3}-7\sqrt {2}?$

$解：長(zhǎng)：?\frac {1}{2}\sqrt {32}=\frac {1}{2}×4\sqrt {2}=2\sqrt {2}?$

$寬：?\frac {1}{3}\sqrt {18}=\frac {1}{3}×3\sqrt {2}=\sqrt {2}?$

$∴周長(zhǎng)為?2(2\sqrt {2}+\sqrt {2})=6\sqrt {2}?$

$對(duì)角線長(zhǎng)度?\sqrt {(2\sqrt {2})2+(\sqrt {2})2}=\sqrt {10}?$

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