第147页 - 亮点给力提优课时作业本八年级数学江苏版

C

$-2 \sqrt {1-x²}$

解：由题意，得$\begin {cases}1010-x≥0\\m +1≥0\\m -8≥0\\8-m≥0\end {cases}$

解得$x≤1010$，$m= 8$

∴$y=3$，$|x-1012|+(\sqrt {1010-x})²$

$=1012-x+1010-x=2022-2x$

又$|x-1012|+(\sqrt {1010-x})²=2024$

∴$2022-2x=2024$

解得$x=-1$

∴$xy=-3$

B

$\sqrt {505}$

解：∵$r≥4$，∴$\frac 1{\sqrt r}+\frac {1}{\sqrt {r+1}}＜1$

∴$a=\frac {1}{r}-\frac {1}{r+1}$

$=( \frac 1{\sqrt {r}}+\frac {1}{\sqrt {r+1}}) (\frac 1{\sqrt {r}}- \frac {1}{\sqrt {r+1}})＜\frac 1{ \sqrt {r}}-\frac {1}{\sqrt {r+1}}$，即$ a ＜ b $

又$ c = \frac {1}{r(\sqrt r+\sqrt {r+1})}= \frac {r+1-r}{r(\sqrt {r}+\sqrt {r+1})} $

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

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

且$ r= \sqrt {r} · \sqrt {r}＜ \sqrt {r} · \sqrt {r+1}$

∴$c＞ \frac {\sqrt {r+1}-\sqrt r}{\sqrt r · \sqrt {r+1}}$

又$ b =\frac 1{ \sqrt r}-\frac 1{\sqrt r+1}$

$=\frac {\sqrt {r+1}-\sqrt r}{\sqrt r · \sqrt {r+1}}$

∴$b＜c$

则$a$，$b$，$c $之间的大小关系为$a＜b＜c$

B

-1

解：∵$\sqrt {2}＞1$，$\sqrt {3}＞ \sqrt {2}$，$\sqrt {4}＞ \sqrt {3}···$，$\sqrt {101}＞ \sqrt {100}$

∴$x =\frac {2}{1+1}+\frac {2}{\sqrt 2+\sqrt {2}}+\frac {2}{\sqrt 3+\sqrt {3}}+···+\frac {2}{\sqrt {100}+\sqrt {100}}＞$

$\frac {2}{1+\sqrt {2}}+\frac {2}{\sqrt 2+\sqrt {3}}+\frac {2}{\sqrt 3+\sqrt {4}}+···+\frac {2}{\sqrt {100}+\sqrt {101}}$

又$\frac {2}{1+\sqrt {2}}+\frac {2}{\sqrt 2+\sqrt {3}}+\frac {2}{\sqrt 3+\sqrt {4}}+···+\frac {2}{\sqrt {100}+\sqrt {101}}$

$=2×(\sqrt {2}-1+ \sqrt {3}- \sqrt {2}+ \sqrt {4}- \sqrt {3}+···+ \sqrt {101}- \sqrt {100})$

$=2(\sqrt {101}-1)$

且$2(\sqrt {101}-1)＞18$

∴$x＞2(\sqrt {101}-1)＞18$，即$x＞18$

同理，得$x=\frac {2}{1+1}+\frac {2}{\sqrt 2+\sqrt {2}}+\frac {2}{\sqrt 3+\sqrt {3}}+···+\frac {2}{\sqrt {100}+\sqrt {100}}＜$

$\frac {2}{1+1}+\frac {2}{\sqrt 2+1}+\frac {2}{\sqrt 3+\sqrt {2}}+···+\frac {2}{\sqrt {100}+\sqrt {99}} $

又$\frac {2}{1+1}+\frac {2}{\sqrt 2+1}+\frac {2}{\sqrt 3+\sqrt {2}}+···+ \frac {2}{\sqrt {100}+\sqrt {99}}$

$=2×(\frac {1}{2}+ \sqrt {2}-1+ \sqrt {3}- \sqrt {2}+···+\sqrt {100}- \sqrt {99})$

$=2×(\frac {1}{2}+ \sqrt {100}-1)=19$

∴$x＜19$

则$18＜x＜19$