f(x,y)=x^2y^2,则f(x^2y.f(x,y)）=

设f(x+y,x-y)=x^2-y^2,则f(x,y)=设f(x+y,x-y)=x^2-y^2,则f(x,y)=设f(x+y,x-y)=x^2-y^2,则f(x,y)=f(x+y，x-y)=x²-y²=(x+y)(x-y

若f(x+y,y/x)=x^2-y^2,则F(X,Y)=若f(x+y,y/x)=x^2-y^2,则F(X,Y)=若f(x+y,y/x)=x^2-y^2,则F(X,Y)=令X=x+y,Y=y/x,解得x=X/(1+Y),y=XY/(1+Y)F

已知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)=已知f(x+y,y/x)=x^2-y^2,则f(x,y)=令m=x+yn=y/x代入f(x+y,y/x)=x^2-y^2得m

f(x+y)=f(x)+f(y)+2xyf(x+y)=f(x)+f(y)+2xyf(x+y)=f(x)+f(y)+2xy游客1:定义在R上的函数f(x)满足f(x+y)=f(x)+f(y)+2xy(x,y∈R)f(1)=2,则f(-3)=?

证f(x+y)+f(x-y)=2f(x)f(y)为偶函数能否取-y?f(x+y)+f(x-y)=2f(x)f(y)取-y即把y换成-yf(x+y)+f(x-y)=2f(x)f(-y)比较两式得f(y)=f(-y)所以为偶函数?那为什么书上不

已知F(X+Y,X-Y)=X^2*Y+X*Y^2,则F(X,Y)=?已知F(X+Y,X-Y)=X^2*Y+X*Y^2,则F(X,Y)=?已知F(X+Y,X-Y)=X^2*Y+X*Y^2,则F(X,Y)=?令m=X+Yn=X-Y有X=(m+n

f(x+Y)+f(x-y)=2f(x)f(Y)求其是偶函数急f(x+Y)+f(x-y)=2f(x)f(Y)求其是偶函数急f(x+Y)+f(x-y)=2f(x)f(Y)求其是偶函数急令y=-x,代入,f(0)+f(2x)=2f(x)f(-x)

求证f（x+y）+f(x-y)=2f(x)f(y)是周期函数求证f（x+y）+f(x-y)=2f(x)f(y)是周期函数求证f（x+y）+f(x-y)=2f(x)f(y)是周期函数首先f(x)=0肯定是解.如果不恒等于0,那么一定有f(x)

f(x+y)+f(x-y)=2f(x)cosy,求证f(x)为周期函数f(x+y)+f(x-y)=2f(x)cosy,求证f(x)为周期函数f(x+y)+f(x-y)=2f(x)cosy,求证f(x)为周期函数取y=π/2,那么f(x+π/

f[(x+y)/2]f[(x+y)/2]f[(x+y)/2]问题问得不完整.这个式子可以理解为：中点的函数值小于两端点函数值的平均值.成立的条件是,f(x)必须是一个下凸函数,比如开口向上的二次函数,如f(x)=x^2；反之,若f(x)是一

f(x)+f(y)=f(x+y)+2x＞2f(x＞2)证明为增函数f(x)+f(y)=f(x+y)+2x＞2f(x＞2)证明为增函数证明f(a²-2a-2)＜3是x＞2和f(x)＞2.这两个分开的f(x)+f(y)=f(x+y)+

f(x+y,x-y)=(x^2-y^2)/2xy,求f(x,y)!f(x+y,x-y)=(x^2-y^2)/2xy,求f(x,y)!f(x+y,x-y)=(x^2-y^2)/2xy,求f(x,y)!1.换元,令x+y=m&x-y=n则x=(

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)f(x+y,y/x)=x^2-y^2求f(x,y)令x+y=t,y/x=u则y=xu代入x+y=t,得x+xu=t(1)u≠-1,即y≠-x

f(x+y,y/x)=x^2-y^2,求f(x,y),详见里面f(x+y,y/x)=x^2-y^2,求f(x,y),详见里面f(x+y,y/x)=x^2-y^2,求f(x,y),详见里面设u=x+y,v=y/x可以解出x=u/(v+1),y

f(x+y,x-y)=2xy(x-y),求f(x,y)f(x+y,x-y)=2xy(x-y),求f(x,y)f(x+y,x-y)=2xy(x-y),求f(x,y)f(x+y,x-y)=(1/2)[(x+y)+(x-y)][(x+y)-(x-

已知函数f(x)满足f(2)=1/2,2f(x)f(y)=f(x+y)+f(x-y),则f(2012)=?已知函数f(x)满足f(2)=1/2,2f(x)f(y)=f(x+y)+f(x-y),则f(2012)=?已知函数f(x)满足f(2)

已知函数f(x)满足f(1)=1/4,f(x)+f(y)=4f(x+y/2)*f(x-y/2)则f(-2011)=?已知函数f(x)满足f(1)=1/4,f(x)+f(y)=4f(x+y/2)*f(x-y/2)则f(-2011)=?已知函数

函数f(x)=x^2-2x,若f(x+1)+f(y+1)≤f(x)+f(y)≤0,则点P(x,y)所形成的区域面积为多少?--各位注意了,不用到其他网页找了,因为没有一模一样的题...函数f(x)=x^2-2x,若f(x+1)+f(y+1)

f(xy,x-y)=x^2+3xy+y^2,则f(x,y)=?f(xy,x-y)=x^2+3xy+y^2,则f(x,y)=?f(xy,x-y)=x^2+3xy+y^2,则f(x,y)=?f(xy,x-y)=x^2+3xy+y^2=(x-y)

设f（x+y,xy）=x^2+y^2,则f（x,y）设f（x+y,xy）=x^2+y^2,则f（x,y）设f（x+y,xy）=x^2+y^2,则f（x,y）f(x+y,xy)=x^2+y^2=(x+y)^2-2xyf(x,y)=x^2-2y