Question

n>0,化简 根号【1+n平方分之一+（n+1)的平方分之一】

Answer 1

1+1/n^2+1/(n+1)^2
=[n^2(n+1)^2+(n+1)^2+n^2]/[n(n+1)]^2
={[n(n+1)]^2+2n(n+1)+1}/[n(n+1)]^2
=(n^2+n+1)^2/[n(n+1)]^2,
n>0,
∴原式=（n^2+n+1)/[n(n+1)].