题目内容
去括号,合并同类项
(1)-3(2s-5)+6s;
(2)3x-[5x-(
x-4)];
(3)6a2-4ab-4(2a2+
ab);
(4)-3(2x2-xy)+4(x2+xy-6)
(1)-3(2s-5)+6s;
(2)3x-[5x-(
1 |
2 |
(3)6a2-4ab-4(2a2+
1 |
2 |
(4)-3(2x2-xy)+4(x2+xy-6)
考点:去括号与添括号,合并同类项
专题:
分析:(1)先去括号,再合并同类项即可;
(2)先去小括号,再去中括号,再合并同类项即可;
(3)先去括号,再合并同类项即可;
(4)先去括号,再合并同类项即可.
(2)先去小括号,再去中括号,再合并同类项即可;
(3)先去括号,再合并同类项即可;
(4)先去括号,再合并同类项即可.
解答:解:(1)-3(2s-5)+6s
=-6s+15+6s
=15;
(2)3x-[5x-(
x-4)]
=3x-[5x-
x+4]
=3x-5x+
x-4
=-
x+4;
(3)6a2-4ab-4(2a2+
ab)
=6a2-4ab-8a2-2ab
=-2a2-6ab;
(4)-3(2x2-xy)+4(x2+xy-6)
=-6x2+3xy+4x2+4xy-24
=-2x2+7xy-24.
=-6s+15+6s
=15;
(2)3x-[5x-(
1 |
2 |
=3x-[5x-
1 |
2 |
=3x-5x+
1 |
2 |
=-
3 |
2 |
(3)6a2-4ab-4(2a2+
1 |
2 |
=6a2-4ab-8a2-2ab
=-2a2-6ab;
(4)-3(2x2-xy)+4(x2+xy-6)
=-6x2+3xy+4x2+4xy-24
=-2x2+7xy-24.
点评:此题考查了整式的运算,用到的知识点是去括号、合并同类项,在去括号时要注意符号的变化和去括号的顺序.
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