人气：433 ℃ 时间：2020-01-27 03:36:38
解答
2a2+3a+2a+1−a2−a−5a+2−3a2−4a−5a−2+2a2−8a+5a−3=（2a+1）+1a+1-（a-3）-1a+2-（3a+2）+1a−2+（2a-2）-1a−3=[（2a+1）-（a-3）-（3a+2）+（2a-2）]+（1a+1-1a+2+1a−2-1a−3）=1a+1-1a+2+1a−2-1a−3=1(a...
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