解
√p-2与q^2-8q+16互为相反数
∴√p-2+(q^2-8q+16)=0
∴√p-2+(q-4)^2=0
∴p-2=0,q-4=0
∴p=2,q=4
∴x^2+y^2-(pxy+q)
=x^2+y^2-(2xy+4)
=(x^2+y^2-2xy)-4
=(x-y)^2-2^2
=(x-y+2)(x-y-2)