零五网 全部参考答案 启东中学作业本 2025年启东中学作业本七年级数学上册苏科版宿迁专版 第20页解析答案
1. 去括号:
(1)$4a - 2(b - 3c) = $
4a-2b+6c
; (2)$-5a + \frac{1}{2}(4x - 6) = $
-5a+2x-3

(3)$3x + [4y - (7z + 3)] = $
3x+4y-7z-3
; (4)$-3a^{3} - [2x^{2} - (5x + 1)] = $
-3a³-2x²+5x+1

答案:(1)4a-2b+6c (2)-5a+2x-3 (3)3x+4y-7z-3 (4)-3a³-2x²+5x+1
解析:
(1)解:$4a - 2(b - 3c)$
$=4a - 2b + 6c$
(2)解:$-5a + \frac{1}{2}(4x - 6)$
$=-5a + 2x - 3$
(3)解:$3x + [4y - (7z + 3)]$
$=3x + (4y - 7z - 3)$
$=3x + 4y - 7z - 3$
(4)解:$-3a^{3} - [2x^{2} - (5x + 1)]$
$=-3a^{3} - (2x^{2} - 5x - 1)$
$=-3a^{3} - 2x^{2} + 5x + 1$
2. 添括号:
(1)$2 - x^{2} + 2xy - y^{2} = 2 -$
$x²-2xy+y²$
; (2)$a - (b - c + d) = a - d +$
$-b+c$

(3)$5x + 3x^{2} - 4y^{2} = 5x -$
$4y²-3x²$
; (4)$-3p + 3q - 1 = 3q -$
$3p+1$

答案:(1)x²-2xy+y² (2)-b+c (3)4y²-3x² (4)3p+1
解析:
(1) $x^{2} - 2xy + y^{2}$
(2) $-b + c$
(3) $4y^{2} - 3x^{2}$
(4) $3p + 1$
3. 先去括号,再合并同类项:
(1)$a - (2a - 2)$; (2)$-(5x + y) - 3(2x - 3y)$;
(3)$2x^{2} - (7 + x) - (3x + 4x^{2})$; (4)$-(3a^{2} - 2a + 1) + (a^{2} - 5a + 7)$;
(5)$a^{2} - 3[a^{2} - 2(a^{2} - a) + 1]$; (6)$3a^{2}b - 2[ab^{2} - 2(a^{2}b - 2ab^{2})]$;
(7)$5a^{2} - [3a - (2a - 3) + 4a^{2}]$; (8)$2(a^{2}b - 3ab^{2}) - 3(2ab^{2} - \frac{5}{6}a^{2}b)$。
答案:(1)-a+2 (2)-11x+8y (3)-2x²-4x-7 (4)-2a²-3a+6 (5)4a²-6a-3 (6)7a²b-10ab² (7)a²-a-3 (8)$\frac{9}{2}a^{2}b-12ab^{2}$
解析:
(1)解:原式$=a - 2a + 2 = -a + 2$
(2)解:原式$= -5x - y - 6x + 9y = -11x + 8y$
(3)解:原式$= 2x^{2} - 7 - x - 3x - 4x^{2} = -2x^{2} - 4x - 7$
(4)解:原式$= -3a^{2} + 2a - 1 + a^{2} - 5a + 7 = -2a^{2} - 3a + 6$
(5)解:原式$= a^{2} - 3[a^{2} - 2a^{2} + 2a + 1] = a^{2} - 3[-a^{2} + 2a + 1] = a^{2} + 3a^{2} - 6a - 3 = 4a^{2} - 6a - 3$
(6)解:原式$= 3a^{2}b - 2[ab^{2} - 2a^{2}b + 4ab^{2}] = 3a^{2}b - 2[5ab^{2} - 2a^{2}b] = 3a^{2}b - 10ab^{2} + 4a^{2}b = 7a^{2}b - 10ab^{2}$
(7)解:原式$= 5a^{2} - [3a - 2a + 3 + 4a^{2}] = 5a^{2} - [a + 3 + 4a^{2}] = 5a^{2} - a - 3 - 4a^{2} = a^{2} - a - 3$
(8)解:原式$= 2a^{2}b - 6ab^{2} - 6ab^{2} + \frac{5}{2}a^{2}b = \frac{9}{2}a^{2}b - 12ab^{2}$
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