零五网 › 全部参考答案› 通城学典课时作业本答案 › 通城学典课时作业本八年级数学苏科版江苏 › 第131页

第131页

信息发布者：

解：原式$=3×\frac {\sqrt {2x}}{4}-2x×\frac {\sqrt {2x}}{x}+\frac {5}{4}×\frac {\sqrt {2x}}{10}$

$ =\frac {3}{4}\sqrt {2x}-2\sqrt {2x}+\frac {1}{8}\sqrt {2x}$

$ =(\frac {3}{4}-2+\frac {1}{8})\sqrt {2x}$

$ =-\frac {9}{8}\sqrt {2x}$

解：原式$=\sqrt {3}-\sqrt {3}+\frac {2(\sqrt {3}+1)}{(\sqrt {3}-1)(\sqrt {3}+1)}-2\sqrt {3}$

$ =0+\sqrt {3}+1-2\sqrt {3}$

$ =1-\sqrt {3}$

 解：$\because x=\sqrt{2}+1，$

 $\therefore x-1=\sqrt{2}，$

 $\therefore (x-1)^2=2，$

即$x^2-2x+1=2，$

 $\therefore x^2-2x=1，$

 $\therefore x^2-2x+2=1+2=3$

 解：$\because x=\frac{\sqrt{5}-1}{2},y=\frac{\sqrt{5}+1}{2}，$

 $\therefore x+y=\frac{\sqrt{5}-1}{2}+\frac{\sqrt{5}+1}{2}=\sqrt{5}，$

 $xy=\frac{\sqrt{5}-1}{2}×\frac{\sqrt{5}+1}{2}=\frac{(\sqrt{5})^2-1^2}{4}=1，$

 $\therefore x^2+xy+y^2=(x+y)^2-xy=(\sqrt{5})^2-1=5-1=4$

 解：$\because 3＜\sqrt{10}＜4，$

 $\therefore -4＜-\sqrt{10}＜-3，$

 $\therefore 2＜6-\sqrt{10}＜3，$

 $\therefore 6-\sqrt{10}$的整数部分$a=2，$小数部分$b=6-\sqrt{10}-2=4-\sqrt{10}，$

 $\therefore (2a+\sqrt{10})b=(4+\sqrt{10})(4-\sqrt{10})=4^2-(\sqrt{10})^2=16-10=6$

$\sqrt{2}$