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第95页

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解：原式$=(3a)²-(2a-b)²$

$=(3a+2a-b)(3a-2a+b)$

$= (5a - b)(a + b)$

解：原式$=(2a-b)²-(6a)²$

$=(2a-b+6a)(2a-b-6a)$

$= -(8a - b)(4a + b)$

解：原式$=(x+3y+x-3y)(x+3y-x+3y)$

$=2x·6y$

$= 12xy$

解:原式​$=(5m-5n)²-(m+n)²$​
​$=(5m-5n-m-n)(5m-5n+m+n)$​
​$=(4m-6n)(6m-4n)$​
​$=4(2m-3n)(3m-2n)$​

解:原式​$=\frac {1000×1000}{(252+248)×(252-248)}$​
​$=\frac {1000×1000}{500×4}$​
​$=500$​

解$：$原式$= (1-\frac {1}{2})× (1+\frac {1}{2})× (1-\frac {1}{3}) × (1+\frac {1}{3})×(1-\frac {1}{4})×(1+\frac {1}{4})×...×(1-\frac {1}{2025})×(1+\frac {1}{2025})$

$=\frac {1}{2}×\frac {3}{2}×\frac {2}{3}×\frac {4}{3}×\frac {3}{4}×\frac {5}{4}×...×\frac {2023}{2024}×\frac {2025}{2024}×\frac {2026}{2025}$

$=\frac {1}{2}×\frac {2026}{2025}$

$=\frac {1013}{2025}$

 设大圆盘的直径为$d_1$cm，小圆盘的直径为$d_2$cm. 根据题意，得$\pi(\frac{d_1}{2})^{2}-4\pi(\frac{d_2}{2})^{2}=7\pi，$即$d_1^{2}-4d_2^{2}=28.$ 分解因式，得$(d_1 + 2d_2)(d_1 - 2d_2)=28.$ $\because d_1,d_2$均为整数，$\therefore d_1 + 2d_2，$$d_1 - 2d_2$均为整数. 又$\because d_1 + 2d_2$和$d_1 - 2d_2$的奇偶性相同，$\therefore \begin{cases}d_1 + 2d_2 = 14\\d_1 - 2d_2 = 2\end{cases}，$解得$\begin{cases}d_1 = 8\\d_2 = 3\end{cases}，$$\therefore \frac{d_1}{2}=4，$$\frac{d_2}{2}=1.5.$ $\therefore$大、小圆盘的半径分别为4cm和1.5cm