第148页 - 亮点给力提优课时作业本八年级数学江苏版

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$解：(1)∵\frac {1}{n\sqrt {n+1}+(n+1)\sqrt {n}}$

$=\frac { \sqrt {n+1}- \sqrt {n}}{[n \sqrt {n+1}+(n+1) \sqrt {n}](\sqrt {n+1}- \sqrt {n})}$

$=\frac { \sqrt {n+1}-\sqrt n }{\sqrt {n(n+1)}}=\frac 1{ \sqrt {n}}- \frac {1}{n+1}$

$∴\frac {1}{\sqrt 2+2}+\frac {1}{2\sqrt {3}+3\sqrt {2}}+\frac {1}{3\sqrt {4}+4\sqrt {3}}+···+ \frac {1}{n\sqrt {n+1}+(n+1)\sqrt {n}} $

$=\frac 1{\sqrt {1}}- \frac {1}{\sqrt 2}+\frac {1}{\sqrt 2}-\frac {1}{\sqrt 3}+···+\frac 1{\sqrt n}-\frac 1{\sqrt {n+1}}$

$=1-\frac {1}{\sqrt {n+1}} $

$则原不等式可化为 \frac {7}{8}＜1-\frac {1}{\sqrt {n+1}}＜\frac {8}{9}$

$解得63＜n＜80$

$又n为正整数$

$∴n的最小值为64，n的最大值为79，$

$即n的最大值与最小值之差为15$

$(2)∵\sqrt x= \sqrt {a}+\frac {1}{\sqrt a}$

$∴x=a+\frac {1}{a}+2$

$∴x-2=a+\frac {1}{a}，即(x-2)²=(a+\frac {1}{a})^2$

$∴x²-4x=a²+\frac {1}{a²}-2=(a-\frac {1}{a})²$

$∴原式=\frac {(x+3)(x-2)}{x} · \frac {x(x-2)}{x+3}-\frac {x-2+ \sqrt {x²-4x}}{ x-2- \sqrt {x²-4x}}$

$=(a+\frac {1}{a})^2-\frac {a+\frac {1}{a}+\sqrt {(a-\frac 1{a})^2}}{a-\frac {1}{a}-\sqrt {(a-\frac 1{a})^2}}$

$又0＜a＜1，∴a-\frac {1}{a}＜0$

$∴原式=(a+\frac {1}{a})^2 -\frac {a+\frac {1}{a}+\frac {1}{a}-a}{a+\frac 1{a}+a-\frac 1{a}} $

$=a²+\frac {1}{a²}+2-\frac {1}{a²}$

$=a²+2$

解：∵$(x+ \sqrt {x²+1})(y+ \sqrt {y²+1})= 1$

∴$x + \sqrt {x²+1} = \frac {1}{y+\sqrt {y²+1}}$

$ =\frac {\sqrt {y²+1}-y}{(y+\sqrt {y²+1})(\sqrt {y²+1}-y)}$

$= \sqrt {y²+1}-y$

∴$x+y= \sqrt {y²+1}- \sqrt {x²+1}$

∴$(x+y)²=(\sqrt {y²+1}- \sqrt {x²+1})²$

即$x^2+y²+2xy=x²+y²- 2 \sqrt {(x²+1)(y²+1)} + 2.$

∴$\sqrt {(x²+1)(y²+1)}=1-xy$，即$(x²+1)(y²+1)=(1-xy)²$

∴$x²y²+x²+y²+1=x²y²+1-2xy$，即$x²+y²+2xy=0$

∴$(x+y)²=0$，则$x+y=0$

解：∵$(x+ \sqrt {x²+1})(y+ \sqrt {y²+1})=1$

∴$(\sqrt {x²+1}-x)(x+ \sqrt {x²+1})(y+\sqrt {y²+1})=\sqrt {x²+1}-x$

$(x+\sqrt {x²+1})(y+\sqrt {y²+1})(\sqrt {y²+1}-y)= \sqrt {y²+1}-y$

即$y+\sqrt {y²+1}= \sqrt {x²+1} -x①$

$x+ \sqrt {x²+1}=\sqrt {y²+1}-y②$

由①+②，得$x+y+ \sqrt {x²+1}+\sqrt {y²+1}$

$= \sqrt {x²+1}+ \sqrt {y²+1}-(x+y)$

即$x+y=-(x+y)$，则$x+y=0$