计算:999...99(2005个9)x999...9(2005个9)+1999...9(2005的9),所得的结果,并说出末尾有几个零.
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计算:999...99(2005个9)x999...9(2005个9)+1999...9(2005的9),所得的结果,并说出末尾有几个零.
计算:999...99(2005个9)x999...9(2005个9)+1999...9(2005的9),所得的结果,并说出末尾有几个零.
计算:999...99(2005个9)x999...9(2005个9)+1999...9(2005的9),所得的结果,并说出末尾有几个零.
设a=2005个9
则a+1=1后面2005个0
则原式=a²+a=a(a+1)
=2005个9×1后面2005个0
=2005个9后面2005个0
999...99(2005个9)x999...9(2005个9)+1999...9(2005个9)
=999...99(2005个9)x[999...9(2005个9)+1]+10000..000(2005个0)
=999...9900000[2005个9,2005个0]+10000..000(2005个0)
=1000...00000(4010个0)
末尾有4010个零
999...99(2005个9)*999...9(2005个9)+1999...9(2005的9)
=[100...00(2005个0)-1]*[100...00(2005个0)-1]+1999...9(2005的9)
=100...000(4010个0)-2*100...00(2005个0)+1+1999...9(2005的9)
=100...000(4010个0)+[-...
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999...99(2005个9)*999...9(2005个9)+1999...9(2005的9)
=[100...00(2005个0)-1]*[100...00(2005个0)-1]+1999...9(2005的9)
=100...000(4010个0)-2*100...00(2005个0)+1+1999...9(2005的9)
=100...000(4010个0)+[-200...00(2005个0)+100...00(2005的9)]
=100...000(4010个0)-100...00(2005个0)
=999...999(2005个9)00...00(2005个0)
所以末尾有2005个0
收起