myworld123456
幼苗
共回答了24个问题采纳率:100% 举报
[(n-1999)+(2000-n)]^2
=(n-1999)^2+(2000-n)^2+2(n-1999)*(2000-n)
=1+2(n-1999)*(2000-n)
[(n-1999)+(2000-n)]^2
=[n-1999+2000-n]^2
=1
(n-1999)*(2000-n)=0
由已知x不是0
x-13+x^-1=0
x+x^-1=13
(x+x^-1)^2
=x^2+2*x*x^-1+x^-2
=169
x^2+x^-2
=169-2*x*x^-1
=169-2
=167
D
1年前
4