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信息发布者：
$3\sqrt{5}+3$
解：​$(1)$​原式​$=2+\sqrt {5}-2-9+3-\sqrt {5}$​
​$=-6$​
解：​$(2)$​原式​$=\sqrt {2}×(\sqrt {2}+\frac {\sqrt {2}}{2})-\frac {3\sqrt {2}-2\sqrt {2}}{\sqrt {2}}$​
​$=\sqrt {2}×\frac {3\sqrt {2}}{2}-\frac {\sqrt {2}}{\sqrt {2}}$​
​$=3-1$​
​$=2$​
解：​$(3)$​原式​$=(3\sqrt {3}×3\sqrt {6}+\frac {4}{5}×5\sqrt {2}-8×\frac {\sqrt {2}}{2})÷\sqrt {2}$​
​$=(27\sqrt {2}+4\sqrt {2}-4\sqrt {2})÷\sqrt {2}$​
​$=27\sqrt {2}÷\sqrt {2}$​
​$=27$​
解：​$(4) $​原式​$=[\sqrt {2}+(\sqrt {3}-\sqrt {5})]×[\sqrt {2}-(\sqrt {3}-\sqrt {5})]$​
​$=(\sqrt {2})^2-(\sqrt {3}-\sqrt {5})^2$​
​$=2-(3-2\sqrt {15}+5)$​
​$=2-8+2\sqrt {15}$​
​$=-6+2\sqrt {15}$​
解：​$ (1) \frac {\sqrt {2}}{2}+\sqrt {8}-\sqrt {18}-4\sqrt {2}$​
​$=\frac {\sqrt {2}}{2}+2\sqrt {2}-3\sqrt {2}-4\sqrt {2}$​
​$=-\frac {9\sqrt {2}}{2}$​
​$ (2) $​因为​$\frac {\sqrt {2}}{2}÷\sqrt {8}×\sqrt {18}□4\sqrt {2}=-\frac {13}{4}\sqrt {2}$​，
​$ $​所以​$\frac {\sqrt {2}}{2}×\frac {1}{2\sqrt {2}}×3\sqrt {2}□4\sqrt {2}=-\frac {13}{4}\sqrt {2}$​，
​$ $​即​$\frac {3\sqrt {2}}{4}□4\sqrt {2}=-\frac {13}{4}\sqrt {2}$​，
​$ $​又因为​$\frac {3\sqrt {2}}{4}-4\sqrt {2}=-\frac {13}{4}\sqrt {2}$​，
所以□内的符号是“-”。
​$ (3) 12-\frac {7\sqrt {2}}{2}$​
​$ A$​
解：​$(1)$​原式​$=\frac {\sqrt {x}\sqrt {y}(\sqrt {x}-\sqrt {y})}{\sqrt {x}\sqrt {y}(\sqrt {x}+\sqrt {y})}-\frac {\sqrt {x}\sqrt {y}(\sqrt {x}+\sqrt {y})}{\sqrt {x}\sqrt {y}(\sqrt {y}-\sqrt {x})}$​
​$ =\frac {\sqrt {x}-\sqrt {y}}{\sqrt {x}+\sqrt {y}}-\frac {\sqrt {x}+\sqrt {y}}{\sqrt {y}-\sqrt {x}}$​
​$ =\frac {(\sqrt {x}-\sqrt {y})^2}{(\sqrt {x}+\sqrt {y})(\sqrt {x}-\sqrt {y})}-\frac {(\sqrt {x}+\sqrt {y})^2}{(\sqrt {y}-\sqrt {x})(\sqrt {y}+\sqrt {x})}$​
​$ =\frac {x+y-2\sqrt {xy}}{x-y}-\frac {x+y+2\sqrt {xy}}{y-x}$​
​$ =\frac {x+y-2\sqrt {xy}+x+y+2\sqrt {xy}}{x-y}$​
​$ =\frac {2(x+y)}{x-y}$​
​$ $​当​$x=3$​，​$y=2$​时，原式​$=\frac {2×(3+2)}{3-2}=10$​
解：​$(2)$​原式​$=(\frac {1}{a-\sqrt {ab}}+\frac {1}{\sqrt {ab}+b})·\frac {(\sqrt {a}+\sqrt {b})(\sqrt {a}-\sqrt {b})}{\sqrt {ab}}$​
​$=[\frac {1}{\sqrt {a}(\sqrt {a}-\sqrt {b})}+\frac {1}{\sqrt {b}(\sqrt {a}+\sqrt {b})}]·\frac {(\sqrt {a}+\sqrt {b})(\sqrt {a}-\sqrt {b})}{\sqrt {ab}}$​
​$=\frac {\sqrt {a}+\sqrt {b}}{a\sqrt {b}}+\frac {\sqrt {a}-\sqrt {b}}{b\sqrt {a}}$​
​$=\frac {\sqrt {ab}+b}{ab}+\frac {a-\sqrt {ab}}{ab}$​
​$=\frac {a+b}{ab}$​
当​$a=\sqrt {3}+1$​，​$b=\sqrt {3}-1$​时，
​$a+b=\sqrt {3}+1+\sqrt {3}-1=2\sqrt {3}$​，
​$ab=(\sqrt {3}+1)(\sqrt {3}-1)=2$​，
所以原式​$=\frac {2\sqrt {3}}{2}=\sqrt {3}$​
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