Question

化简 根号下n(n+1)(n+2)(n+3)

Answer 1

错了吧
应该是根号下[n(n+1)(n+2)(n+3)+1]
n(n+1)(n+2)(n+3)+1
=[n(n+3)][(n+1)(n+2)]+1
=(n^2+3n)[(n^2+3n)+2]+1
=(n^2+3n)^2+2(n^2+3n)+1
=(n^2+3n+1)^2
所以根号下[n(n+1)(n+2)(n+3)+1]
=绝对值(n^2+3n+1)
若n^2+3n+1<0
即(-3-√5)/2
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Answer 2

不会打平方的符号．．把n的平方打成了n2.．．．n2+3n的平方打成了（n2+3n)2
应该看的懂吧
先移项n(n+3)(n+1)(n+2)＋1
前两项相乘后两项相乘得：(n2+3n)(n2+3n+2)+1
再把（n2+3n)看成一个整体．（n2+3n)2＋2（n2+3n)＋1
最后配方．．．再开方．．