Question

第八题怎么做

Answer 1

证：
²ⁿ√[n(n+1)(n+2)(n+3)+1]
=²ⁿ√[n(n+3)(n+1)(n+2)+1]
=²ⁿ√[(n²+3n)(n²+3n+2)+1]
=²ⁿ√[(n²+3n)²+2(n²+3n)+1]
=²ⁿ√[(n²+3n+1)²]
=ⁿ√|n²+3n+1|
n(n+1)(n+2)(n+3)+1能够开平方，因此²ⁿ√[n(n+1)(n+2)(n+3)+1]不是最简根式。

Answer 2

如为非最简根式
令n(n+1)(n+2)(n+3)+1=m^2
有
n(n+1)(n+2)(n+3)=m^2-1=(m-1)(m+1)
[n(n+3)][(n+1)(n+2)]=(m-1)(m+1)
(n^2-3n)(n^2-3n+2)=(m-1)(m+1)
n^2-3n=m-1
m=n^2-3n+1
原式=m＝ⁿ√(n²+3n+1)
所以原式为非最简。