已知a,b,m,n满足am-bn=3,am+bm=4,求[(a+b)^2+(a-b)^2][(m-n)^2+2mn]的值

[(a+b)^2+(a-b)^2][(m-n)^2+2mn]
=（a^2+b^2+2ab+a^2+b^2-2ab)(m^2+n^2-2mn+2mn)
=2(a^2+b^2)(m^2+n^2)
=2[(am)^2+(bn)^2+(an)^2+(bm)^2]
题目写错了一个字母,应为an+bm=4
分别将两式平方相加
25=（am-bn)^2+(an+bm)^2
=(am)^2+(bn)^2-2abmn+(an)^2+(bm)^2+2abmn
=(am)^2+(bn)^2+(an)^2+(bm)^2
原式=25*2=50