解由(根号a +2),/b一4/均为非负数
且(根号a +2)+/b一4/=0
故a+2=0且b-4=0
解得a=-2,b=4
故(x^2+y^2)一(axy+b)
=(x^2+y^2)一(-2xy+4)
=x^2+2xy+y^2-4
=(x+y)^2-4
=（x+y)^2-2^2
=(x+y+2)(x+y-2)