第124页 - 苏科版八年级（初二）数学学习与评价答案（上下册）

A

$解：原式=\sqrt {40×10}$

$ =\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}$

$解：将a=2，b=-8，c=5代入得：$

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

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

$将a=2+\sqrt {3}，b=2-\sqrt {3}代入原式得：$

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

$解：菱形的边长为\sqrt {96}÷4=\sqrt {6}(\mathrm {cm})$

$∵∠DAB = 120°$

$∴∠DAO= 60°$

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

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

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

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

$∴a²+\frac {1}{a²}=8$

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

$∴a-\frac {1}{a}=±\sqrt {6}$