第109页 - 苏科版八年级数学课课练答案（上下册）

$解：矩形的另外一条边长为2\sqrt {6}÷\sqrt {2}=2\sqrt {3}\mathrm {cm}$

$ 对角线的长为\sqrt {(\sqrt {2})²+(2\sqrt {3})²}=\sqrt {14}\mathrm {cm}$

$ \sqrt{3}$

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

$解：(2)①原式=\frac {2(\sqrt{5}－\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}－\sqrt{3})}$

$=\sqrt{5}－\sqrt{3}$

$②原式=\frac {5－3}{\sqrt{5}+\sqrt{3}}$

$=\frac {(\sqrt{5})²－(\sqrt{3})²}{\sqrt{5}+\sqrt{3}}$

$=\frac {(\sqrt{5}+\sqrt{3})(\sqrt{5}－\sqrt{3})}{\sqrt{5}+\sqrt{3}}$

$=\sqrt{5}－\sqrt{3}$

$(3)原式=2(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+...+\sqrt{2n+1}-\sqrt{2n-1})$

$=2(\sqrt{2n+1}-1)$

$ \frac {\sqrt {3}}{3}$

$\sqrt{6}$

$ \frac {\sqrt {2}}{2}$

$ \frac{\sqrt{6}}{2}$

$\frac{\sqrt{6}}{3}$

$\frac{\sqrt{6}}{6}$