《2025年暑假总动员七年级数学沪科版合肥工业大学出版社》

1. 简便运算:
(1)$48^{2}+48×24+12^{2};$
(2)$101×99-99.5^{2};$
(3)$75×2.6^{2}-12×3.5^{2}.$

答案： 解:
(1) 原式 $ = (48 + 12)^2 = 3600 $.
(2) 原式 $ = (100 + 1) × (100 - 1) - (100 - \frac{1}{2})^2 $
$ = 100^2 - 1^2 - (100^2 - 100 + \frac{1}{4}) $
$ = 100^2 - 1 - 100^2 + 100 - \frac{1}{4} = 98\frac{3}{4} $.
(3) 原式 $ = 3 × 5^2 × 2.6^2 - 3 × 2^2 × 3.5^2 = 3 × [(5 × 2.6)^2 - (2 × 3.5)^2] = 3 × [(13 + 7) × (13 - 7)] = 3 × 20 × 6 = 360 $.

2. 计算:
(1)$(a-2b+3)(a+2b+3);$
(2)$(x+2y+1)(x-2y+1)-(x-2y+1)^{2};$
(3)$(a+b-c+d)(a-b+c+d).$

答案： 解:
(1) 原式 $ = (a + 3 - 2b)(a + 3 + 2b) $
$ = (a + 3)^2 - (2b)^2 $
$ = a^2 + 6a + 9 - 4b^2 $.
(2) 原式 $ = (x + 1)^2 - 4y^2 - (x - 2y + 1)^2 $
$ = (x + 1)^2 - 4y^2 - [(x + 1)^2 - 4y(x + 1) + 4y^2] $
$ = -8y^2 + 4xy + 4y $.
(3) 原式 $ = [(a + d) + (b - c)][(a + d) - (b - c)] $
$ = (a + d)^2 - (b - c)^2 $
$ = a^2 + 2ad + d^2 - b^2 + 2bc - c^2 $.

3. 计算:
(1)$6(7+1)(7^{2}+1)(7^{4}+1)(7^{8}+1)+1;$
(2)$(3^{8}+1)(3^{4}+1)(3^{2}+1)(3+1).$

答案： 解:
(1) 原式 $ = (7 - 1)(7 + 1)(7^2 + 1)(7^4 + 1)(7^8 + 1) + 1 $
$ = (7^2 - 1)(7^2 + 1)(7^4 + 1)(7^8 + 1) + 1 $
$ = (7^4 - 1)(7^4 + 1)(7^8 + 1) + 1 $
$ = (7^8 - 1)(7^8 + 1) + 1 $
$ = 7^{16} - 1 + 1 = 7^{16} $.
(2) 原式 $ = \frac{(3^8 + 1)(3^4 + 1)(3^2 + 1)(3 + 1)(3 - 1)}{2} $
$ = \frac{(3^8 + 1)(3^4 + 1)(3^2 + 1)(3^2 - 1)}{2} $
$ = \frac{(3^8 + 1)(3^4 + 1)(3^4 - 1)}{2} $
$ = \frac{(3^8 + 1)(3^8 - 1)}{2} = \frac{3^{16} - 1}{2} $.

4. (1)已知$a^{2}+b^{2}=13,ab=6$,求$(a+b)^{2},(a-b)^{2}$的值;
(2)已知$(a+b)^{2}=7,(a-b)^{2}=4$,求$a^{2}+b^{2},ab$的值;
(3)已知$a(a-1)-(a^{2}-b)=2$,求$\frac {a^{2}+b^{2}}{2}-ab$的值;
(4)已知$(2027+m)(2025+m)=n$,试求$(2027+m)^{2}+(2025+m)^{2}$的值.(用含n的代数式表示)

答案： 解:
(1) 因为 $ a^2 + b^2 = 13 $, $ ab = 6 $, 所以 $ (a + b)^2 = a^2 + b^2 + 2ab = 13 + 2 × 6 = 25 $, $ (a - b)^2 = a^2 + b^2 - 2ab = 13 - 2 × 6 = 1 $.
(2) 因为 $ (a + b)^2 = 7 $, $ (a - b)^2 = 4 $, 所以 $ a^2 + 2ab + b^2 = 7 \cdots $ ①, $ a^2 - 2ab + b^2 = 4 \cdots $ ②,
① + ②, 得 $ 2(a^2 + b^2) = 11 $, 即 $ a^2 + b^2 = \frac{11}{2} $.
① - ②, 得 $ 4ab = 3 $, 即 $ ab = \frac{3}{4} $.
(3) 由 $ a(a - 1) - (a^2 - b) = 2 $, 得 $ a - b = -2 $.
所以 $ \frac{a^2 + b^2}{2} - ab = \frac{1}{2}(a^2 + b^2 - 2ab) = \frac{1}{2}(a - b)^2 = \frac{1}{2} × (-2)^2 = 2 $.