1002^2-1002*4+4
问题描述：
1002^2-1002*4+4
896^2+208*896+104^2
(181^2-61^2)/(301^2-181^2)
已知x+(1/x)=-3,求x^2+(1/x^2)的值．
答
1002^2-1002*4+4 =(1002-2)^2=1000000896^2+208*896+104^2=(896+104)^2=1000000 (181^2-61^2)/(301^2-181^2)=(181+61)(181-61)/(301+181)(301-181)=242x120/482x120=121/241x+(1/x)=-3,两边平方,x^2+(1/x^2)+2=9x^2+...