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$2+\sqrt{5}$
$6\sqrt{15}$
$ \begin{aligned} 解:原式&=-\sqrt {60}×\sqrt {96} \\ &=-\sqrt {10×576} \\ &=-24 \sqrt{10} \\ \end{aligned}$
$ \begin{aligned} 解:原式&=2\sqrt {35×\frac {10}{7}} \\ &=\sqrt {100×2} \\ &=10\sqrt{2} \\ \end{aligned}$
$ \begin{aligned}解:原式&=\sqrt {\frac {8}{5}}×(-\sqrt {30}) \\ &=-\sqrt {16×3} \\ &=-4\sqrt{3} \\ \end{aligned}$
$ \begin{aligned}解:原式&=\sqrt {36m^3n^5} \\ &=\sqrt {36m^2n^4•mn} \\ &=6mn^2\sqrt{mn} \\ \end{aligned}$
$ \begin{aligned} 解:原式&=\sqrt {42}×\sqrt {-24} \\ &=-\sqrt {7×144} \\ &=-12\sqrt{7} \\ \end{aligned}$
$ \begin{aligned}解:原式&=4\sqrt {ab}•\sqrt {a(a^2-2ab+b^2)} \\ &=4\sqrt {ab}•\sqrt {a(a-b)^2} \\ &=4\sqrt {a^2(a-b)b} \\ &=4a(a-b) \sqrt{b} \\ \end{aligned}$
解: 原式​$=\frac {(a - b)^2}{3(a - b)}\div \frac {b - a}{ab}=\frac {a - b}{3}·\frac {ab}{-(a - b)}=-\frac {1}{3}ab$​，
当​$a = \sqrt {2} + 1$​，​$b = \sqrt {2} - 1$​时，
原式​$=-\frac {1}{3}(\sqrt {2} + 1)(\sqrt {2} - 1)=-\frac {1}{3}[(\sqrt {2})^2 - 1^2]=-\frac {1}{3}×(2 - 1)=-\frac {1}{3}$​。
$\sqrt{6\times8 + 1}=7$

$解: \sqrt{(n+1)(n+3)+1}=n+2， $
$ \begin{aligned}证明: \sqrt{(n+1)(n+3)+1}&= \sqrt{n^2+3n+n+3+1} \\ &=\sqrt{n^2+4n+4} \\ &=\sqrt{(n+2)^2} \\ &=n+2. \\ \end{aligned}$
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