第142页 - 亮点给力提优课时作业本八年级数学江苏版

A

$\sqrt {7}$

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

$\frac {3\sqrt {2}+2\sqrt {3}}{6}$

解：∵$ \frac {1}{\sqrt 3-\sqrt {2}} ＜ x ＜ \frac {2}{\sqrt 6-\sqrt {5}}$

∴$\frac {\sqrt {3}+\sqrt {2}}{(\sqrt {3}-\sqrt {2})(\sqrt {3}+\sqrt {2})}＜x＜\frac {2(\sqrt {6}+\sqrt {5})}{(\sqrt {6}-\sqrt {5})(\sqrt {6}+\sqrt {5})}$，即$ \sqrt {3}+ \sqrt {2}＜x＜2 \sqrt {6}+2 \sqrt {5}$

又$3²＜(\sqrt {3}+\sqrt {2})²＜4²$，$9²＜(2 \sqrt {6}+2 \sqrt {5})²＜10²$，∴$3＜x＜10$

则整数$x=4$，$5$，$6$，$7$，$8$，$9$，其和为$4+5+6+7+8+9=39$

$=\frac { \sqrt 2-\sqrt 3}{ \sqrt {6}(\sqrt 2-\sqrt 3) }$

$ =\frac {1}{\sqrt 6}$

$=\frac {\sqrt {6}}{6}$

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

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

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

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

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

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

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

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

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

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

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

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

解：∵$\frac {1}{\sqrt {n+1}+\sqrt {n}}= \sqrt {n+1}- \sqrt {n}(n $为正整数$)$

∴原式$=[(\sqrt {2}-1)+(\sqrt {3}- \sqrt {2})+ \sqrt {4}-\sqrt {3})+···+(\sqrt {2025}- \sqrt {2024})]×(\sqrt {2025}+1)$

$=(\sqrt {2025}-1)×(\sqrt {2025}+1)$

$=2024$

解：∵$13+ \sqrt {48}=1+2×1×2 \sqrt {3}+(2 \sqrt {3})²= (1+2 \sqrt {3})²$

∴$5-(1+2 \sqrt {3})=4-2 \sqrt {3}=(\sqrt {3}-1)²$

∴$3+(\sqrt {3}-1)=2+ \sqrt {3}=\frac {4+2\sqrt {3}}{2}=\frac {\sqrt {3}+1)^2}{2}=(\frac {\sqrt 6+\sqrt 2}2)^2 $

则原式$=\frac {\frac {\sqrt 6+\sqrt 2}2}{\sqrt 6+\sqrt 2}= \frac {1}{2}$