题目

计算: (1) [a(a2b2-ab)-b(-a3b-a2)]÷a2b; (2) . 答案: 解:[a(a2b2-ab)-b(-a3b-a2)]÷a2b =(a3b2-a2b+a3b2+a2b)÷a2b =2a3b2÷a2b =2ab 解: (−12)−2+(2018−π)0−|−4| =4+1-4 =1.
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