初三数学题,化简

√[n*(n+1)*(n+2)*(n+3)+1]化简它

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2009-06-04

0 0

√[n*(n+1)*(n+2)*(n+3)+1] 只要把n*(n+1)*(n+2)*(n+3)+1化成平方数即可　n*(n+1)*(n+2)*(n+3)+1 ＝n(n+3)(n+1)(n+2)+1 ＝(n²+3n)(n²+3n+2)+1 ＝(n²+3n)²＋2(n²+3n)＋1 ＝(n²+3n+1)² 原式＝n²+3n+1。

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终***

2009-06-04

44 0

=√[(n+1)*(n+2)*n*(n+3)+1] =√[(n^2+3n+2)*(n^2+3n)+1] =√[(n^2+3n)^2+2(n^2+3n)+1] =√(n^2+3n+1)^2 =|n^2+3n+1|

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初***

2009-06-04

56 0

暂时根号不考虑，电脑上打麻烦 n(n+1)(n+2)(n+3)+1 =n(n+3)*[(n+1)(n+3-1)]+1 =n(n+3)[n(n+3)+(n+3)-n-1]+1 =n(n+3)[n(n+3)+2]+1 =n(n+3)^2+2*n(n+3)+1 =[n(n+3)+1]^2 现在考虑根号 =n(n+3)+1 别忘记加绝对值，

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