Question

kuai,本人急用!第二题可以用全等三角形的知识

Answer 1

1,n^2+3n+1=(n+2)*n+n+2-1

=(n+2)(n+1)-1

n^2+3n+1=n(n+3)+1

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

=n(n+1)(n+2)(n+3)-n^2-3n+n^2+3n+2-1

=n(n+1)(n+2)(n+3)+1

x=y

2,无图