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$?9\sqrt{2}?$

$解:原式?=3\sqrt{6}+2\sqrt{6}-9×\frac {\sqrt{6}}{18}?$

$?=3\sqrt{6}+2\sqrt{6}-\frac {\sqrt{6}}{2}?$

$?=\frac {9\sqrt{6}}{2}$

$解:原式?=10x2\sqrt{xy}×(5÷15)×\sqrt{\frac {y}{x}÷\frac {x}{y}}?$

$?=10x2\sqrt{xy}×\frac {1}{3}×\frac {y}{x}?$

$?=\frac {10}{3}xy\sqrt{xy}$

$解:原式?=3\sqrt{2}+3\sqrt{5}-2\sqrt{5}-5\sqrt{2}?$

$?=\sqrt{5}-2\sqrt{2}?$

$解:原式?=(4+4\sqrt{10}+10)(14-4\sqrt{10})?$

$?=(14+4\sqrt{10})×(14-4\sqrt{10})?$

$?=142-(4\sqrt{10})2?$

$?=196-160?$

$?=36?$

$解:原式?=\frac {1+x+1-x}{x+1}×\frac {(x+1)2}{2(x-1)}?$

$?=\frac {2}{x+1}×\frac {(x+1)2}{2(x-1)}?$

$?=\frac {x+1}{x-1}?$

$將?x=\sqrt{2}+1?代入原式$

$原式?=\frac {\sqrt{2}+1+1}{\sqrt{2}+1-1}=1+\sqrt{2}$

$解:三角形的面積?=\frac {1}{2}×(3-\sqrt{2})×(3+\sqrt{2})=\frac {7}{2}{cm}^2；?$

$三角形的斜邊長?=\sqrt{{(3-\sqrt{2})}^2+{(3+\sqrt{2})}^2}=\sqrt{22}，?$

$∴三角形的周長?=(3-\sqrt{2})+(3+\sqrt{2})+\sqrt{22}=(6+\sqrt{22})\ \mathrm {cm}.?$

$解：?(1)?如圖①所示：$

$?AC+CE=\sqrt{{x}^2+25}+\sqrt{{x}^2-16x+65}，?$

$當?A、??C、??E?在同一直線上，?AC+CE?最?。?

$?(2)?作點?N?關于?x?軸的對稱點?N'，?連接?MN'?交?x?軸于點?P，?此時?PM+PN?的值最小，等于?MN'，?$

$過點?M?作?y?軸的垂線交射線?N'N?于點?A，?如圖②所示.$

$?∵N(3，??2)，?$

$?∴N'(3，??-2).?$

$設直線?MN'?得解析式為?y=kx+b，?$

$則?\{\begin{array}{l}{b=4}\\{3k+b=-2}\end{array}，?$

$解得?\{\begin{array}{l}{k=-2}\\{b=4}\end{array}.?$

$?∴y=-2x+4.?$

$當?-2x+4=0?時，?x=2，?$

$?∴P(2，??0).?$

$在?Rt△AMN'?中，?AM=3，??AN'=6，?$

$?∴MN'=\sqrt{A{M}^2+AN{'}^2}=\sqrt{{3}^2+{6}^2}=3\sqrt{5}.?$

$?∴PM+PN?最小值為?3\sqrt{5}.?$

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