999···9(1997个9)*999···9(1997个9)+1999···9(1997个9)的末尾有多少个零
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![999···9(1997个9)*999···9(1997个9)+1999···9(1997个9)的末尾有多少个零](/uploads/image/z/8598574-46-4.jpg?t=999%C2%B7%C2%B7%C2%B79%EF%BC%881997%E4%B8%AA9%EF%BC%89%2A999%C2%B7%C2%B7%C2%B79%EF%BC%881997%E4%B8%AA9%EF%BC%89%2B1999%C2%B7%C2%B7%C2%B79%EF%BC%881997%E4%B8%AA9%EF%BC%89%E7%9A%84%E6%9C%AB%E5%B0%BE%E6%9C%89%E5%A4%9A%E5%B0%91%E4%B8%AA%E9%9B%B6)
999···9(1997个9)*999···9(1997个9)+1999···9(1997个9)的末尾有多少个零
999···9(1997个9)*999···9(1997个9)+1999···9(1997个9)的末尾有多少个零
999···9(1997个9)*999···9(1997个9)+1999···9(1997个9)的末尾有多少个零
999···9(1997个9)*999···9(1997个9)+1999···9(1997个9)
=999···9(1997个9)*(999···9(1997个9)+1)
=999···9(1997个9)*100000.0000(共1997的0)
所以末尾有1997的零
999···9(1997个9)*999···9(1997个9)+1999···9(1997个9)
=(100..0(1997个0)-1)*(100..0(1997个0)-1)+1999···9(1997个9)
=100..0(1997个0)的平方- 2*100..0(1997个0)+1+1999···9(1997个9)
=100..0(1997+1997个0)- 200.....
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999···9(1997个9)*999···9(1997个9)+1999···9(1997个9)
=(100..0(1997个0)-1)*(100..0(1997个0)-1)+1999···9(1997个9)
=100..0(1997个0)的平方- 2*100..0(1997个0)+1+1999···9(1997个9)
=100..0(1997+1997个0)- 200..0(1997个0)+200..0(1997个0)
=100..0(1997+1997=3994个0)
.......
那么明显有问题的回答你都采纳。。。服了你了
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