题目内容
化简
(1)3x2+2xy-4y2-(3xy-4y2+3x2)
(2)-2(4a2b-5ab2)-2(3a2b-2ab2)
(1)3x2+2xy-4y2-(3xy-4y2+3x2)
(2)-2(4a2b-5ab2)-2(3a2b-2ab2)
分析:(1)先按照去括号法则去掉整式中的小括号,再合并整式中的同类项即可;
(2)先去括号,再合并同类项即可.
(2)先去括号,再合并同类项即可.
解答:解:(1)3x2+2xy-4y2-(3xy-4y2+3x2)
=3x2+2xy-4y2-3xy+4y2-3x2
=-xy;
(2)-2(4a2b-5ab2)-2(3a2b-2ab2)
=-8a2b+10ab2-6a2b+4ab2
=-14a2b+14ab2.
=3x2+2xy-4y2-3xy+4y2-3x2
=-xy;
(2)-2(4a2b-5ab2)-2(3a2b-2ab2)
=-8a2b+10ab2-6a2b+4ab2
=-14a2b+14ab2.
点评:本题考查了整式的加减、去括号法则两个考点.解决此类题目的关键是熟记去括号法则,熟练运用合并同类项的法则,这是各地中考的常考点.
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