第111页 - 苏科版八年级（初二）数学学习与评价答案（上下册）

$解：原式=\sqrt {16}×\sqrt {49}$

$ =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$

$解：∵点O是菱形ABCD的两对角线交点$

$∴AC⊥BD$

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

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

$∴C_{菱形ABCD}=4AB= 8\sqrt {7}$

$解：原式=\sqrt {100}×\sqrt {0.1}$

$ =\sqrt {100×0.1}$

$ =\sqrt {10}$

$解：原式=\sqrt {9×\frac {1}{3}}$

$ =\sqrt {3}$