《2025年名校课堂八年级数学上册北师大版》

1.(2024·安徽)我国古代数学家张衡将圆周率取值为$\sqrt {10}$,祖冲之给出圆周率的一种分数形式的近似值为$\frac {22}{7}$.比较大小:$\sqrt {10}$＞$\frac {22}{7}$.(填“＞”或“＜”)

答案： ＞

2.比较大小:$\sqrt [3]{20}$＜3.(填“＞”或“＜”)

答案： ＜

3.兰生复旦校本经典题 综合实践活动课上,老师给出定理:对于任意两个正数a,b,若$a＞b$,则$\sqrt {a}＞\sqrt {b}$.随后讲解了一道例题:试比较$2\sqrt {3}和3\sqrt {2}$的大小.解:$(2\sqrt {3})^{2}= 12,(3\sqrt {2})^{2}= 18,$$\because 12＜18,$$\therefore 2\sqrt {3}＜3\sqrt {2}.$参考例题的解法,解答下列问题:(1)比较$-3\sqrt {5}和-5\sqrt {3}$的大小.解：$(3\sqrt{5})^{2}=$45，$(5\sqrt{3})^{2}=$75，$\because$45＜75，$\therefore 3\sqrt{5}＜5\sqrt{3}$。$\therefore$$-3\sqrt{5}＞-5\sqrt{3}$。(2)比较$\sqrt {6}+\sqrt {2}和\sqrt {5}+\sqrt {3}$的大小.解：$(\sqrt{6}+\sqrt{2})^{2}=$$8+4\sqrt{3}$，$(\sqrt{5}+\sqrt{3})^{2}=$$8+2\sqrt{15}$，$\because (4\sqrt{3})^{2}=$48，$(2\sqrt{15})^{2}=$60，48＜60，$\therefore 4\sqrt{3}＜2\sqrt{15}$。$\therefore$$8+4\sqrt{3}＜8+2\sqrt{15}$。$\therefore$$\sqrt{6}+\sqrt{2}＜\sqrt{5}+\sqrt{3}$。

答案： 解：
(1) $(3\sqrt{5})^{2}=45$，$(5\sqrt{3})^{2}=75$，$\because 45＜75$，$\therefore 3\sqrt{5}＜5\sqrt{3}$。$\therefore -3\sqrt{5}＞-5\sqrt{3}$。
(2) $(\sqrt{6}+\sqrt{2})^{2}=8+4\sqrt{3}$，$(\sqrt{5}+\sqrt{3})^{2}=8+2\sqrt{15}$，$\because (4\sqrt{3})^{2}=48$，$(2\sqrt{15})^{2}=60$，$48＜60$，$\therefore 4\sqrt{3}＜2\sqrt{15}$。$\therefore 8+4\sqrt{3}＜8+2\sqrt{15}$。$\therefore \sqrt{6}+\sqrt{2}＜\sqrt{5}+\sqrt{3}$。

4.课堂上,老师出了一道题:比较$\frac {\sqrt {19}-2}{3}与\frac {2}{3}$的大小.
小明的解法如下:
解:$\frac {\sqrt {19}-2}{3}-\frac {2}{3}= \frac {\sqrt {19}-2-2}{3}= \frac {\sqrt {19}-4}{3}.$
$\because 19＞16,\therefore \sqrt {19}＞4.\therefore \sqrt {19}-4＞0.$
$\therefore \frac {\sqrt {19}-4}{3}＞0.\therefore \frac {\sqrt {19}-2}{3}＞\frac {2}{3}.$
我们把这种比较大小的方法称为作差法.
请利用上述方法比较实数$\frac {\sqrt {94}-3}{9}与\frac {2}{3}$的大小.
解：$\frac{\sqrt{94}-3}{9}-\frac{2}{3}=\frac{\sqrt{94}-3}{9}-\frac{6}{9}=\frac{\sqrt{94}-9}{9}$。$\because 94＞81$，$\therefore \sqrt{94}＞9$。$\therefore \sqrt{94}-9＞0$。$\therefore \frac{\sqrt{94}-9}{9}＞0$。$\therefore \frac{\sqrt{94}-3}{9}＞\frac{2}{3}$。

答案： 解：$\frac{\sqrt{94}-3}{9}-\frac{2}{3}=\frac{\sqrt{94}-3}{9}-\frac{6}{9}=\frac{\sqrt{94}-9}{9}$。$\because 94＞81$，$\therefore \sqrt{94}＞9$。$\therefore \sqrt{94}-9＞0$。$\therefore \frac{\sqrt{94}-9}{9}＞0$。$\therefore \frac{\sqrt{94}-3}{9}＞\frac{2}{3}$。

5.用作商法比较大小:$\frac {\sqrt {7}}{\sqrt {5}}与\frac {\sqrt {5}}{2}$.

答案： 解：$\frac{\sqrt{7}}{\sqrt{5}}=\frac{2\sqrt{7}}{5}$，$\because (2\sqrt{7})^{2}=28$，$5^{2}=25$，$\therefore 2\sqrt{7}＞5$。$\therefore \frac{2\sqrt{7}}{5}＞1$。$\therefore \frac{\sqrt{7}}{\sqrt{5}}＞\frac{\sqrt{5}}{2}$。

6.新考向 阅读理解认真阅读下列解答过程:比较$2-\sqrt {3}与\sqrt {3}-\sqrt {2}$的大小.
解:$\because \frac {1}{2-\sqrt {3}}= \frac {2+\sqrt {3}}{(2-\sqrt {3})(2+\sqrt {3})}= 2+\sqrt {3},$$\frac {1}{\sqrt {3}-\sqrt {2}}= \frac {\sqrt {3}+\sqrt {2}}{(\sqrt {3}-\sqrt {2})(\sqrt {3}+\sqrt {2})}= \sqrt {3}+\sqrt {2},$
又$\because 2＞\sqrt {2},$
$\therefore 2+\sqrt {3}＞\sqrt {3}+\sqrt {2}.\therefore \frac {1}{2-\sqrt {3}}＞\frac {1}{\sqrt {3}-\sqrt {2}}.$$\therefore 2-\sqrt {3}＜\sqrt {3}-\sqrt {2}.$
请仿照上述方法比较$\sqrt {6}-\sqrt {5}与\sqrt {5}-2$的大小.
解：$\because \frac{1}{\sqrt{6}-\sqrt{5}}=\frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6}-\sqrt{5})(\sqrt{6}+\sqrt{5})}=$$\sqrt{6}+\sqrt{5}$，$\frac{1}{\sqrt{5}-2}=\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}=$$\sqrt{5}+2$，又$\because$$\sqrt{6}＞2$，$\therefore$$\sqrt{6}+\sqrt{5}＞\sqrt{5}+2$。$\therefore$$\frac{1}{\sqrt{6}-\sqrt{5}}＞\frac{1}{\sqrt{5}-2}$。$\therefore$$\sqrt{6}-\sqrt{5}＜\sqrt{5}-2$。

答案： 解：$\because \frac{1}{\sqrt{6}-\sqrt{5}}=\frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6}-\sqrt{5})(\sqrt{6}+\sqrt{5})}=\sqrt{6}+\sqrt{5}$，$\frac{1}{\sqrt{5}-2}=\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}=\sqrt{5}+2$，又$\because \sqrt{6}＞2$，$\therefore \sqrt{6}+\sqrt{5}＞\sqrt{5}+2$。$\therefore \frac{1}{\sqrt{6}-\sqrt{5}}＞\frac{1}{\sqrt{5}-2}$。$\therefore \sqrt{6}-\sqrt{5}＜\sqrt{5}-2$。