f(x,y)=2(x-y)+x^2-y^2这种形式的函数算隐函数吗?

f(x+y,x-y)=(x^2-y^2)/2xy,求f(x,y)! 设f(x+y,x-y)=x^2-y^2,则f(x,y)= 若f(x+y,y/x)=x^2-y^2,则F(X,Y)= 已知f(x+y,y/x)=x^2-y^2,则f(x,y)= f(x+y,y/x)=x^2-y^2求f(x,y) f(x+y,y/x)=x^2-y^2,求f(x,y),详见里面 f(x+y,x-y)=2xy(x-y),求f(x,y) f(xy,x-y)=x^2+y^2+xy;f(x,y)/∂x;f(x,y)/∂y (X,Y) f(x,y)={12y^2,0 x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2 f(x+y)=f(x)+f(y)+2xy f(x^2-y^2,x-y)=x+y+x^2-xy,求f'(x,y)对y求导 已知F(X+Y,X-Y)=X^2*Y+X*Y^2,则F(X,Y)=? 设f(x+y,x-y)=x^2+y^2-x*y求f(x,y) f[(x+y)/2] f(x+Y)+f(x-y)=2f(x)f(Y) 求其是偶函数 急 求证f（x+y）+f(x-y)=2f(x)f(y)是周期函数 设f(x+y,x-y)=x^2+xy,求f(x,y)