f(x y,y除以x)=x乘以x减y乘以y,求解f(x,y)

由X+Y=3XY可得X+Y-XY=2XY,用X+Y-XY替换分母中的2XY,所以(2X+2Y-XY)/(X+Y+2XY)=(2X+2Y-XY)/(2X+2Y-XY)=1

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,

原式=-3xy^2*144y^2/x^2*x^3/216y^6=-2*1/y^2=-2/y^2

(x-y)²=(x+y)²-4xy=36+8=44x-y=±√44=±2√11于是（x^4-y^4）除以x^2+y^2/xy乘以（6x+6y）=（x²+y²)(

式子是这样么(2x+y)/(x^2-2xy+y^2）=(2x+y)/(x-y)^2x-3y=0x=3y代入原式(2*3y+y)/(3y-y)^2=7/4yy不能等于0

XY(XY-X^2)(X-Y)=X^2Y(X-Y)(X-Y)=(X-Y)(X^2Y-1)再问：答案是。-X2Y.我不知道过程是如何来的。能告诉我过程吗。谢谢。负X的平方Y

对.前提是x不等于y

.（x-4根号xy+4y）除以（根号x-2根号y）+（2根号xy+x）乘以根号x分之一=(根号x-2根号Y)^2/(根号X-2根号Y)+(2根号XY+X)*根号X/X=根号X-2根号Y+(2X根号Y+

〔(x-2y）/(x^2+xy）〕*〔（x^2-y^2)/（x^2-4xy+4y^2)/x〕=〔(x-2y）/x(x+y）〕*〔（x+y)(x-y)/（x-2y)^2/x〕=〔(x-2y）/x(x+y

=(xy-x的平方）除以(x的平方-2xy+y的平方\xy)乘以(x-y\x的平方)=-x（x-y）×xy/（x-y）²×（x-y）/x²=-y

证明令x=x/y,y=y∵f(xy)=f(x)+f(y)∴f（x/y*y）=f(x/y)+f(y)f(x)=f(x/y)+f(y)∴f(x/y)=f(x)-f(y)

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

挺好的题f(xy)=xf(y)+yf(x)---(1)设y=c=常量则：f(cx)=cf(x)+f(c)x两边求导数f'(cx)*c=cf'(x)+f(c)cf'(cx)-cf'(x)=f(c)此式对

[3x分之2-（x+y分之2）乘以[3x分之x+（y-x^2-xy）]除以x分之x-y{2/3x-2/(x+y)*[x/3x+(y-x²-xy)]}÷(x-y)/x题目错了吧?再问：{3x分

(x/x+y+2y/x+y)乘以xy/x+2y除以(1/x+1/y)=(x+2y)/(x+y)*xy/(x+2y)/[(x+y)/(xy)]=xy/(x+y)*xy/(x+y)=(xy)^2/(x+y

x^2+4y^2-2x+4y+5=0==>(x-1)^2+(2y+1)^3+3=0无解?还是要用到虚数什么的

(1)f(x)=3^xf(y)=3^yf(x+y)=3^(x+y)f(x)*f(y)=3^x*3^y=f(x+y)(2)f(x)=3^xf(y)=3^yf(x)/f(y)=3^x/3^y=3^(x-y

（xy-x²）÷xy分之x²-2xy+y²乘x²分之x-y=-x(x-y)×xy/(x-y)²×(x-y)/x²=-y∴结果与x的值没有关系

x^2+4y^2-4x+4y=5=0?如果条件是x^2+4y^2-4x+4y+5=0,那么变形就有x^2-4x+4+4y^2+4y+1=0,即（x-2）^2+(2y+1)^2=0,很显然x=2,y=-

原式=5x根号(xy)/3根号(y/x)*1/3根号(x/y)=5/9*根号(xy*x/y*x/y)=5x/9y根号(xy)