f(2x y) g(x,xy)

解2(x²-xy)-3(2x²-3xy)=(2x²-6x²)+(-2xy+9xy)=-4x²+7xy再问：����|-15|-8�£�-2��+4*�

1=(x-z-2ab)/xy2=(a²-2ab+b²)/a-b=(a-b)²/a-b=a-

因为f（x+y,xy）=x^2+y^2=（x+y）^2-2xy所以f（x,y）=x^2-2y现对x求导得到：fx(x,y)=2x再对y求导得到：fxy(x,y)=0.所以无论x,y为何值,fxy(x,

f(x,y)={xy/[2-√(4+xy)]=-2-√（4+xy),xy≠0；{4,xy=0,在点(0,0),(1,0)处不连续,在（1,2）处连续.再问：能简述下原因么？再答：f(0+,0+)=-4

先对x求偏导u'x=f'(x,xy,xyz)+yf'(x,xy,xyz)+yzf'(x,xy,xyz)所以u'xy=yf''(x,xy,xyz)+xzf''(x,xy,xyz)+f''(x,xy,xy

设u=xy,v=lnx+g(xy),则x(∂z/∂x)-y(∂z/∂y)=∂f/∂v.原因如下：dz=(∂f/

1.2(Xy+Xy)-3(Xy-xy)-4Xy=2*2xy-0-4xy=4xy-4xy=02.1/2ab-5aC-(3acb)+(3aC-4aC)=1/2ab-5ac-3acb-ac=1/2ab-6a

对.前提是x不等于y

【x²+xy/（x-y）】/【xy/（x-y）】=【x²（x-y）/（x-y）+xy/（x-y）】/【xy/（x-y）】={【x²（x-y）+xy】/（x-y）}/【xy

设a=xy,b=x+y.f(xy,x+y)=x^2+y^2+2xy-2xy=(x+y)^2-2xy把a,b带f(a,b)=b^2-2a所以f(x,y)=y^2-2x同理f(x+y,xy)=x^2+y^

Dz/Dx=2f'+g1+yg2,DDz/DxDy=-2f"+yg12+y^2*g22.

当x=y=0时f(0+0)+f(0)=2f(0)f(0)f(0)²=f(0)f(0)=1或者f(0)=0当y=0时f(x)+f(0)=2f(x)f(0)若f(0)=0f(x)=0若f（0)=

-3(2x^2-xy)+4(x^2+xy-6)=-6x^2+3xy+4x^2+4xy-24=-2x^2+7xy-24再问：thankyou~

(3x^2+2xy)/xy-(2x^2-xy)/xy=(3x^2+2xy-2x^2+xy)/xy=(x^2+3xy)/xy=x(x+3y)/xy=(x+3y)/y

-xy^2+2xy-x=-x(y^2-2y+1)=-x(y-1)^2如果本题有什么不明白可以追问,如果满意请点击“选为满意答案”

1、g'(x)=(x^2)'=2xf’[g'(x)]?=f'(2x)=(2^2x)'=2^2x*ln2*(2x)'=2^(2x+1)*ln22、xy^2-e^xy+3=0（y^2+x*2y*y')-e

定义在R+上的函数f(x)满足f(x)+f(y)+2xy(xy)=f(xy)/f(x+y)对任意x,y∈R+,恒成立,则f（2）=____设x=y=2,则有2f(2)+32=f(4)/f(4)=1,故

x,y都是未知数,你也可以把他们当做t,r那么就是求f（t,r）首先由题意知2x+y=t,2y+x=r用t,r表示x,y,可得x=1/3（2t-r）,y=1/3（2r-t）并将其代入f(2x+y,2y

(x-y)/x²÷(x²-2xy+y²)/xy×(xy-x²)=(x-y)/x²÷(x+y)²/xy×x(y-x)=(x-y)²/

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