题目内容
计算:2(3+1)(32+1)(34+1)(38+1)(316+1)(332+1)+1.
考点:乘方
专题:计算问题(巧算速算)
分析:首先把2(3+1)化成32-1,然后根据平方差公式,求出算式2(3+1)(32+1)(34+1)(38+1)(316+1)(332+1)+1的值是多少即可.
解答:
解:2(3+1)(32+1)(34+1)(38+1)(316+1)(332+1)+1
=8(32+1)(34+1)(38+1)(316+1)(332+1)+1
=(32-1)(32+1)(34+1)(38+1)(316+1)(332+1)+1
=(34-1)(34+1)(38+1)(316+1)(332+1)+1
=(38-1)(38+1)(316+1)(332+1)+1
=(316-1)(316+1)(332+1)+1
=(332-1)(332+1)+1
=364-1+1
=364
=8(32+1)(34+1)(38+1)(316+1)(332+1)+1
=(32-1)(32+1)(34+1)(38+1)(316+1)(332+1)+1
=(34-1)(34+1)(38+1)(316+1)(332+1)+1
=(38-1)(38+1)(316+1)(332+1)+1
=(316-1)(316+1)(332+1)+1
=(332-1)(332+1)+1
=364-1+1
=364
点评:此题主要考查了乘方问题,解答此题的关键是熟练掌握平方差公式.
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