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第143页

信息发布者：

$ D$

$2\sqrt{3}$

$-\sqrt{6}$

$ C$

$\sqrt{6}$

-2

解：∵$a=\frac {\sqrt {m+1}-\sqrt {m}}{\sqrt {m+1}+\sqrt {m}}， b=\frac {\sqrt {m+1}+\sqrt {m}}{\sqrt {m+1}-\sqrt {m}}$，

$ $分母有理化得$a=(\sqrt {m+1}-\sqrt {m})^2$，

$b=(\sqrt {m+1}+\sqrt {m})^2$，

$ ab=1$，

∴$a+b=(\sqrt {m+1}-\sqrt {m})^2+(\sqrt {m+1}+\sqrt {m})^2$

$=4m+2.$

∵$a+b+3ab=2025$，

∴$4m+2+3=2025$，

$ $解得$m=505.$

解$：=\frac {(\sqrt {x}-\sqrt {y})(\sqrt {x}+\sqrt {y})}{\sqrt {x}-\sqrt {y}}$

$= \sqrt {x}+\sqrt {y}$

解$：=\frac {(2+\sqrt {3})²}{2+\sqrt {3}}$

$= 2+\sqrt {3}$

解$：(3)$∵$x＞2，$

∴$\frac {x-2+\sqrt {x^2-4}}{x+2+\sqrt {x^2-4}}$

$=\frac {(\sqrt {x-2})^2+\sqrt {(x+2)(x-2)}}{(\sqrt {x+2})^2+\sqrt {(x+2)(x-2)}}$

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

$=\frac {\sqrt {x-2}}{\sqrt {x+2}}$

$=\frac {\sqrt {x^2-4}}{x+2}$

解$：\frac {\sqrt {2}+2\sqrt {3}+\sqrt {5}}{(\sqrt {2}+\sqrt {3})×(\sqrt {3}+\sqrt {5})}$

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

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

$ = \frac {1}{\sqrt {3}+\sqrt {5}} +\frac {1}{\sqrt {2}+\sqrt {3}}$

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

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

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

$=\frac {\sqrt {10}-\sqrt {6}-\sqrt {15}+5}{\sqrt {10}+2}$

$=\frac {(\sqrt {10}-\sqrt {6}-\sqrt {15}+5)×(\sqrt {10}-2)}{(\sqrt {10}+2)×(\sqrt {10}-2)}$

$=\frac {3\sqrt {10}-3\sqrt {6}}{6}$

$=\frac {\sqrt {10}-\sqrt {6}}{2}.$