怎样化简根号下n(n+1)(n+2)(n+3)
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怎样化简根号下n(n+1)(n+2)(n+3)
怎样化简根号下n(n+1)(n+2)(n+3)
怎样化简根号下n(n+1)(n+2)(n+3)
我来回答,错了吧\x0d
应该是根号下[n(n+1)(n+2)(n+3)+1]\x0d
\x0d
n(n+1)(n+2)(n+3)+1\x0d
=[n(n+3)][(n+1)(n+2)]+1\x0d
=(n^2+3n)[(n^2+3n)+2]+1\x0d
=(n^2+3n)^2+2(n^2+3n)+1\x0d
=(n^2+3n+1)^2\x0d
所以根号下[n(n+1)(n+2)(n+3)+1]\x0d
=绝对值(n^2+3n+1)\x0d
\x0d
\x0d
若n^2+3n+1<0\x0d
即(-3-√5)/2<n<(-3+√5)/2\x0d
则原式=-(n^2+3n+1)=-n^2-3n-1\x0d
\x0d
若n^2+3n+1≥0\x0d
即n≤(-3-√5)/2,n≥(-3+√5)/2\x0d
则原式=n^2+3n+1 19451希望对你有帮助!
14,23分别拆开
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