《2025年课堂点睛八年级数学下册人教版安徽专版》

1.计算:
(1)$\sqrt{45}+\sqrt{108}+\sqrt{1\frac{1}{3}}-\sqrt{125}+2\sqrt{5}$;
(2)$|-\sqrt{2}|+(\sqrt{2}-\frac{1}{2})^{2}-(\sqrt{2}+\frac{1}{2})^{2}$;
(3)$(2\sqrt{3}-1)(2\sqrt{3}+1)-(1 - 2\sqrt{3})^{2}$;

答案：
(1)解:原式$=3\sqrt{5}+6\sqrt{3}+\frac{2\sqrt{3}}{3}-5\sqrt{5}+2\sqrt{5}=\frac{20\sqrt{3}}{3}-2\sqrt{5}+2\sqrt{5}=\frac{20\sqrt{3}}{3}$.
(2)解:原式$=\sqrt{2}+(\sqrt{2}-\frac{1}{2}+\sqrt{2}+\frac{1}{2})(\sqrt{2}-\frac{1}{2}-\sqrt{2}-\frac{1}{2})=\sqrt{2}+2\sqrt{2}\times(-1)=-\sqrt{2}$.
(3)解:原式$=12 - 1-(1 - 4\sqrt{3}+12)=11 - 13 + 4\sqrt{3}=4\sqrt{3}-2$.

2.先化简,再求值:$(\frac{a + b}{a - b})^{2}\cdot\frac{2a - 2b}{3a + 3b}-\frac{4a^{2}}{a^{2}-b^{2}}\div\frac{3a}{b}$,其中$a = \sqrt{3},b = \sqrt{2}$.

答案： 解:原式$=\frac{(a + b)^2}{(a - b)^2}\cdot\frac{2(a - b)}{3(a + b)}-\frac{4a^2}{(a + b)(a - b)}\cdot\frac{b}{3a}=\frac{2(a + b)}{3(a - b)}-\frac{4ab}{3(a + b)(a - b)}=\frac{2(a + b)^2-4ab}{3(a + b)(a - b)}=\frac{2(a^2 + b^2)}{3(a + b)(a - b)}$,当$a=\sqrt{3},b=\sqrt{2}$时,原式$=\frac{2\times(3 + 2)}{3(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\frac{10}{3}$.

3.小李同学在计算$\sqrt{45}-\sqrt{15}\times(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}})$时,解题过程如下:
解:$\sqrt{45}-\sqrt{15}\times(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}})$
$=\sqrt{3^{2}\times5}-\sqrt{15}\times(\frac{\sqrt{5}}{\sqrt{15}}-\frac{\sqrt{3}}{\sqrt{15}})$
$=3\sqrt{5}-\sqrt{15}\times\frac{\sqrt{5}-\sqrt{3}}{\sqrt{15}}$
$=3\sqrt{5}-(\sqrt{5}-\sqrt{3})$
$=3\sqrt{5}-\sqrt{5}+\sqrt{3}$
$=2\sqrt{5}+\sqrt{3}$
(1)在以上解题步骤中用到了______(多选).
A.等式的基本性质
B.二次根式的性质:$\sqrt{a^{2}} = a(a\geq0)$
C.二次根式的乘法法则:$\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}$
D.通分
(2)小李同学算完后,发现式子的计算方法很复杂,请你写出一种简便的方法计算此题.

答案：
(1)BD
(2)解:简便方法如下:$\sqrt{45}-\sqrt{15}\times(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}})=\sqrt{3^2\times5}-(\sqrt{15}\times\frac{1}{\sqrt{3}}-\sqrt{15}\times\frac{1}{\sqrt{5}})=3\sqrt{5}-(\sqrt{5}-\sqrt{3})=3\sqrt{5}-\sqrt{5}+\sqrt{3}=2\sqrt{5}+\sqrt{3}$.