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第40页

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解：∵$1≤a≤2$，∴$0≤a-1≤1$，即$0≤ \sqrt {a-1}≤1$

∴$\sqrt {a-1}+1＞0$，$\sqrt {a-1}-1≤0$

∴$m= \sqrt {a+2\sqrt {a-1}}+ \sqrt {a-2\sqrt {a-1}}$

$=\sqrt {a-1+2\sqrt {a-1}+1}+ \sqrt {a-1-2\sqrt {a-1}+1}$

$=\sqrt {a-1}+1+1- \sqrt {a-1}$

$=2$

∴$\sqrt {\mathrm {m^3}+1}=\sqrt {9}=3$

解：原式$= \frac {\sqrt {x²+y²}(x- \sqrt {x²+y²}) }{ (x+ \sqrt {x²+y²})(x- \sqrt {x²+y²})}+\frac { \sqrt {x²+y²}(x+\sqrt {x²+y²})}{(x-\sqrt {x^2+y^2})(x+\sqrt {x^2+y^2})} +\frac {2\sqrt {x²+y²}+xy}{y²}$

$=\frac {2x\sqrt {x²+y²}}{-y}+\frac {2x\sqrt {x²+y²}+xy}{y²}=\frac {x}y$

又$x=\sqrt {3} -\sqrt {2}$，$y=\sqrt {3}+\sqrt {2}$

∴原式$=\frac {\sqrt {3}-\sqrt 2}{\sqrt 3+\sqrt 2}=5- 2\sqrt {6}$

$解：在Rt△ABC中，∠ACB=90°，AC=3，BC= 3\sqrt {2}$

$∴S_{△ABC}=\frac {1}{2}AC · BC=\frac {9\sqrt {2}}{2}，AB=\sqrt {AC²+BC²}=3 \sqrt {3}$

$∵D是AB的中点$

$∴CD=\frac {1}{2}\ \mathrm {AB}=\frac {3\sqrt {3}}{2}，S_{△BCD}=\frac {1}{2}\ \mathrm {S}_{△ABC}=\frac {9\sqrt {2}}{4}$

$又BE⊥CD$

$∴S_{△BCD}=\frac {1}{2}CD · BE $

$即BE=\frac {2S_{△BCD}}{CD}=\frac {2×\frac {9\sqrt {2}}{4}}{\frac {3\sqrt 3}2} =\sqrt {6}$