题目内容
化简
(1)3a2-2a-3+4a2-7a+5;
(2)2(2a-3b)+3(2b-3a);
(3)2a2-[
(ab-a2)+8ab]-
ab;
(4)8x2-[-3x-(2x2-7x-5)+3]+4x.
(1)3a2-2a-3+4a2-7a+5;
(2)2(2a-3b)+3(2b-3a);
(3)2a2-[
1 |
2 |
1 |
2 |
(4)8x2-[-3x-(2x2-7x-5)+3]+4x.
分析:(1)直接合并同类项即可;
(2)先去括号,再合并同类项即可;
(3)、(4)先去小括号,再去中括号,最后合并同类项即可.
(2)先去括号,再合并同类项即可;
(3)、(4)先去小括号,再去中括号,最后合并同类项即可.
解答:解:(1)原式=7a2-9a+2.
(2)原式=4a-6b+6b-9a
=-5a;
(3)原式=2a2-[
ab-
a2+8ab]-
ab
=2a2-
ab+
a2-8ab-
ab
=
a2-9ab;
(4)原式=8x2-[-3x-2x2+7x+5+3]+4x
=8x2+3x+2x2-7x-5-3+4x
=6x2-8.
(2)原式=4a-6b+6b-9a
=-5a;
(3)原式=2a2-[
1 |
2 |
1 |
2 |
1 |
2 |
=2a2-
1 |
2 |
1 |
2 |
1 |
2 |
=
5 |
2 |
(4)原式=8x2-[-3x-2x2+7x+5+3]+4x
=8x2+3x+2x2-7x-5-3+4x
=6x2-8.
点评:本题考查的是整式的加减,熟知整式的加减实质上就是合并同类项是解答此题的关键.

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