试题

题目:
计算:
(1)(x-2)(x+1)-x(x-1);
(2)(x-1)(4-2x)(3-2x);
(3)(2x-3)(3x+2)-2(x+1)(x一1);
(4)[ab(1-a)-2a(b-
1
2
)]·(2a3b2).
答案
解:(1)(x-2)(x+1)-x(x-1)
=x2-x-2-x2+x
=-2.

(2)(x-1)(4-2x)(3-2x)
=(-2x2+6x-4)(-2x+3)
=4x3-6x2-12x2+18x+8x-12
=4x3-18x2+26x-12.

(3)(2x-3)(3x+2)-2(x+1)(x一1)
=6x2+4x-9x-6-2x2+2
=4x2-5x-4.

(4)[ab(1-a)-2a(b-
1
2
)]·(2a3b2
=[ab-a2b-2ab+a]·(2a3b2
=[-a2b-ab+a]·(2a3b2
=-2a5b3-2a4b3+2a4b2
解:(1)(x-2)(x+1)-x(x-1)
=x2-x-2-x2+x
=-2.

(2)(x-1)(4-2x)(3-2x)
=(-2x2+6x-4)(-2x+3)
=4x3-6x2-12x2+18x+8x-12
=4x3-18x2+26x-12.

(3)(2x-3)(3x+2)-2(x+1)(x一1)
=6x2+4x-9x-6-2x2+2
=4x2-5x-4.

(4)[ab(1-a)-2a(b-
1
2
)]·(2a3b2
=[ab-a2b-2ab+a]·(2a3b2
=[-a2b-ab+a]·(2a3b2
=-2a5b3-2a4b3+2a4b2
考点梳理
整式的混合运算.
(1)先算乘法,再合并同类项即可.
(2)先算乘法,再算乘法,最后合并即可.
(3)先算乘法,再合并即可.
(4)先算括号内的乘法,再合并,最后算乘法即可.
本题考查了整式的混合运算的应用,主要考查学生的计算能力,注意运算顺序.
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