已知a向量=[(x+ay)i+yj]/(x+y)^2为某一二元函数的梯度,则a等于

设此二元函数为F(x,y),则Fx=(x+ay)/(x+y)^2,Fy=y/(x+y)^2Fxy=[a(x+y)^2-2(x+y)*(x+ay)]/(x+y)^4=[(a-2)x-ay]/(x+y)^3;Fyx=[(a-2)x-ay]=-2y/(x+y)^3;由Fxy=Fyx,得a-2=0 （1）-a=-2 （2）右（1）（2）两式解得a=2....