第113页 - 苏科版八年级（初二）数学学习与评价答案（上下册）

C

$解：原式=\sqrt {100×3}$

$ =10\sqrt {3}$

$解：原式=\sqrt {5×3×4×5}$

$ =10\sqrt {3}$

$解：原式=2b\sqrt {a}$

$解：原式=\sqrt {a²(a+2)}$

$=a\sqrt {a+2}$

$解：原式=-3ab\sqrt {b}$

$解：原式=\sqrt {(2n-2)²}$

$ =2n-2$

$解：(1)原式=\sqrt {10×\frac {2}{5}}=\sqrt {4}=2$

$(2)原式=-\sqrt {\frac {2}{3}×48}=-\sqrt {32}=-4\sqrt {2}$

$(3)原式=\sqrt {ab×\frac {b³}{a}}=\sqrt {b^4}=b²$

$(4)原式=a²×\frac {1}{a}×\sqrt {2b^5×8b}=a\sqrt {16b^6}=4ab³$

$解：(1)原式=\sqrt {3ab×a}=a\sqrt {3b}$

$将a=\frac {1}{2}，b=3代入得原式=\frac {1}{2}×\sqrt {3×3}=\frac {3}{2}$

$(2)原式=\sqrt {ab³×2b}=b²\sqrt {2a}$

$将a=\frac {1}{2}，b=3代入原式=3²×\sqrt {2×\frac {1}{2}}=9$

$解：V=2\sqrt {6}×\sqrt {3}×\sqrt {2}=2\sqrt {6×3×2}=12(\mathrm {cm}³)$

$解：(1)2\sqrt {12}=\sqrt {2^2×12}=\sqrt {48}$

$∵48＞27$

$∴2\sqrt {12}＞\sqrt {27}$

$(2)2\sqrt {3}=\sqrt {2^2×3}=\sqrt {12}，3\sqrt {2}=\sqrt {3^2×2}=\sqrt {18}$

$∵12＜18$

$∴2\sqrt {3}＜3\sqrt {2}$

$解：\sqrt {3}×\sqrt {\frac {1}{3}}=1$