题目内容
计算:
(1)(4ay2-3a)+(2+4a-ay2)-(2ay2+a);
(2)3(x2-2xy)-2[(x2-2y)+(x2-4y)].
(1)(4ay2-3a)+(2+4a-ay2)-(2ay2+a);
(2)3(x2-2xy)-2[(x2-2y)+(x2-4y)].
分析:(1)、(2)先去括号,再合并同类项即可.
解答:解:(1)原式=4ay2-3a+2+4a-ay2-2ay2-a
=(4-1-2)ay2-(3-4+1)a+2
=ay2+2;
(2)原式=3(x2-2xy)-2(x2-2y)-2(x2-4y)
=3x2-6xy-2x2+4y-2x2+8y
=-x2-6xy+4y.
=(4-1-2)ay2-(3-4+1)a+2
=ay2+2;
(2)原式=3(x2-2xy)-2(x2-2y)-2(x2-4y)
=3x2-6xy-2x2+4y-2x2+8y
=-x2-6xy+4y.
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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