Question

已知:m^2+n^2=5,x^2+y^2=401,求多项式(mx+ny)^2+(nx-my)^2的值

Answer 1

(mx+ny)^2+(nx-my)^2=m^2(x^2)+2mnxy+n^2(y^2)+n^2(x^2)-2mnxy+m^2(y^2)=

（x^2+y^2）*(m^2+n^2)=401*5=2005 应该是这样哈,有很久没有算了,作个参考吧