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第143頁(yè)

信息發(fā)布者：

D

$2\sqrt {3}$

$-\sqrt {6}$

C

$\sqrt {6}$

-2

$解:分母有理化得a=(\sqrt{m+1}-\sqrt{m})^{2}$

$b=(\sqrt{m+1}+\sqrt{m})^{2},ab=1$

$∴a+b=(\sqrt{m+1}-\sqrt{m})^{2}+(\sqrt{m+1}+\sqrt{m})^{2}=4m+2$

$∵a+b+3ab=2025，∴4m+2+3=2025，∴m=505$

$解:∵ x＞2,∴原式=\frac{(\sqrt{x-2})^{2}+\sqrt {(x+2)(x-2)}}{(\sqrt{x+2})^{2}+\sqrt {(x+2)(x-2)}}$

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

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

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

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

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

$解:(1)原式=\frac {(\sqrt {x}+\sqrt {y})(\sqrt {x}-\sqrt {y})}{\sqrt {x}-\sqrt {y}}=\sqrt {x}+\sqrt {y}$

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

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

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

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

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