《2025年课堂点睛八年级数学上册北师大版》

1. 先化简,再求值.
(1)$\frac {4}{3}\sqrt {9a^{3}}+\sqrt {\frac {a}{4}}-a\sqrt {16a}$,其中$a= 2;$
(2)$2(a+\sqrt {3})(a-\sqrt {3})-a(a-\sqrt {2})+6$,其中$a= \sqrt {2}-1;$

答案： （1）解：原式$=\frac{4}{3}× 3a\sqrt{a}+\frac{\sqrt{a}}{2}-4a\sqrt{a}=4a\sqrt{a}+\frac{\sqrt{a}}{2}-4a\sqrt{a}=\frac{\sqrt{a}}{2}$，当$a=2$时，原式$=\frac{\sqrt{2}}{2}$．（2）解：原式$=2(a^{2}-3)-a^{2}+\sqrt{2}a+6=2a^{2}-6-a^{2}+\sqrt{2}a+6=a^{2}+\sqrt{2}a$，当$a=\sqrt{2}-1$时，原式$=(\sqrt{2}-1)^{2}+\sqrt{2}(\sqrt{2}-1)=3-2\sqrt{2}+2-\sqrt{2}=5-3\sqrt{2}$．

2. 已知$\sqrt {a-2027}+|b+2028|= 0$,求$\sqrt {(a+b)^{2}}$的值.

答案： 解：$\because \sqrt{a-2027}+\vert b+2028\vert =0$，$\therefore a-2027=0$，$b+2028=0$，$\therefore a=2027$，$b=-2028$，$\therefore \sqrt{(a+b)^{2}}=\sqrt{(2027-2028)^{2}}=1$．

3. 已知实数$x,y满足x-2\sqrt {x-3}+\sqrt {y-2}= 2$,求$\frac {\sqrt {x}}{\sqrt {x+y}}$的值.

答案： 解：$\because x-2\sqrt{x-3}+\sqrt{y-2}=2$，$\therefore x-3-2\sqrt{x-3}+1+\sqrt{y-2}=0$，$\therefore (\sqrt{x-3}-1)^{2}+\sqrt{y-2}=0$，$\therefore \sqrt{x-3}-1=0$，$\sqrt{y-2}=0$，解得$x=4$，$y=2$，则$\frac{\sqrt{x}}{\sqrt{x+y}}=\frac{\sqrt{4}}{\sqrt{4+2}}=\frac{\sqrt{6}}{3}$．

4. 已知$a= 2+\sqrt {3},b= 2-\sqrt {3}.$
(1)求$ab= $1;
(2)求$a^{2}+b^{2}-ab= $13.

答案： （1）1 （2）13

5. 在解决问题“已知$a= \frac {1}{\sqrt {2}-1}$,求$3a^{2}-6a-1$的值”时,小明是这样分析与解答的:
$\because a= \frac {1}{\sqrt {2}-1}= \frac {\sqrt {2}+1}{(\sqrt {2}-1)(\sqrt {2}+1)}= \sqrt {2}+1,$
$\therefore a-1= \sqrt {2},$
$\therefore (a-1)^{2}= 2,a^{2}-2a+1= 2.$
$\therefore a^{2}-2a= 1.$
$\therefore 3a^{2}-6a= 3,3a^{2}-6a-1= 2.$
请你根据小明的分析过程,解决如下问题:
若$a= \frac {2}{3-\sqrt {7}}$,求$2a^{2}-12a+1$的值.

答案： 解：$\because a=\frac{2}{3-\sqrt{7}}=\frac{2(3+\sqrt{7})}{(3-\sqrt{7})(3+\sqrt{7})}=\frac{2(3+\sqrt{7})}{2}=3+\sqrt{7}$，$\therefore a-3=\sqrt{7}$，$\therefore (a-3)^{2}=7$．即$a^{2}-6a+9=7$．$\therefore a^{2}-6a=-2$．$\therefore 2a^{2}-12a=-4$．$\therefore 2a^{2}-12a+1=-4+1=-3$．即$2a^{2}-12a+1$的值为$-3$．