第115页 - 苏科版八年级（初二）数学学习与评价答案（上下册）

$解：原式=\frac {\sqrt {3}}{2}$

$解：原式=\frac {3}{x}$

$解：原式=\frac {4b²}{a}$

$解：原式=\frac {2\sqrt {2}}{3}$

$解：原式=\sqrt {\frac {25}{9}}$

$ =\frac {5}{3}$

$解：原式=\frac {7×2}{9}$

$ =\frac {14}{9}$

$解：原式=\frac {x\sqrt {x}}{5y²}$

$解：原式=-\sqrt {\frac {54}{3}}$

$ =-\sqrt {18}$

$ =-3\sqrt {2}$

解：原式$=-\frac {1}{3}\sqrt {\frac {2}{98}}$

$ =-\frac {1}{3}×\frac {1}{7}$

$ =-\frac {1}{21}$

$解：原式=\sqrt {\frac {16}{5}÷\frac {8}{5}}$

$ =\sqrt {2}$

解：原式$=\frac {1}{2}×\sqrt {6ab÷3a}$

$ =\frac {1}{2}\sqrt {2b}$

$\sqrt {\frac {125}{26}}$

$5\sqrt {\frac {5}{26}}$

$解：(2)\sqrt {n-\frac {n}{n^2+1}}=n\sqrt {\frac {n}{n^2+1}}$

$证明：\sqrt {n-\frac {n}{n^2+1}}=\sqrt {\frac {n(n^2+1)-n}{n^2+1}}=\sqrt {\frac {n^3+n-n}{n^2+1}}$

$=\sqrt {\frac {n^3}{n^2+1}}=n \sqrt {\frac {n}{n^2+1}}$

解：设下底长为$x$

$(2+x)×\sqrt {2}×\frac {1}{2}=\sqrt {50}$

$ x=8$

∴梯形的下底长为$8$