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電子課本網(wǎng) › 第111頁

第111頁

信息發(fā)布者：

$解：原式?=\sqrt {16}×\sqrt {49}?$

$? =4×7?$

$? =28?$

$解：原式?=\sqrt {144}×\sqrt {9}?$

$? =12×3?$

$? =36?$

$解：原式?=\sqrt {25}×\sqrt {2a}?$

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

$解：原式?=\sqrt {9a2b2}\sqrt {a}?$

$? =3ab\sqrt {a}?$

$解：?\sqrt {32}×2\sqrt {2}=2\sqrt {64}=16(\mathrm {cm}2)?$

$解：原式?=2\sqrt {3}×3\sqrt {3}?$

$? =6×3?$

$? =18?$

$解：原式?=-2×\sqrt {400}?$

$? =-2×20?$

$? =-40?$

$解：原式?=\sqrt {20a×\frac {a3}{5}}?$

$? =\sqrt {4a^4}?$

$? =2a2?$

$解：原式?=\sqrt {0.4}×\sqrt {0.4}×\sqrt {9}?$

$? =0.4×3?$

$? =1.2?$

$解：原式?=2\sqrt {2}×\sqrt {7}×\sqrt {7}×\sqrt {2}?$

$? =4×7?$

$? =28 ?$

$解：原式?=\sqrt {x^4y2}?$

$? =x2y?$

$解：∵點?O?是菱形?ABCD?的兩對角線交點$

$∴?AC⊥BD?$

$?OA =\frac {1}{2}AC= 2\sqrt {3}，??OB=\frac {1}{2}BD= 4?$

$∴?AB=\sqrt {OA2+OB2}=\sqrt {28}=2\sqrt {7}?$

$∴?C_{菱形ABCD}=4AB= 8\sqrt {7}?$

$解：原式?=\sqrt {100}×\sqrt {0.1}?$

$? =\sqrt {100×0.1}?$

$? =\sqrt {10}?$

$解：原式?=\sqrt {9×\frac {1}{3}}?$

$? =\sqrt {3}?$

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