已知f(x y,x-y)=2xy y^2,则f(x,y)=
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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
x-xy=8(1)xy-y=-9(2)则有(1)-(2):X-XY-XY+Y=X+Y-2XY=8-(-9)=17
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,
-xy(x^2y^5-xy^3-y)=-(xy^2)^3+(xy^2)^2+xy^2=-(-2)^3+4-2=8+4-2=10
因为x-y=4xy所以x-2xy-y=2xy2x+3xy-2y=11xyx-2xy-y分之2x+3xy-2y=5.5
再问:二式到三式看不懂,为什么要把1换成y,能不能讲解一下原理。为什么要这样做,为什么不能直接代入?再答:第二行:把x换成y/x,把y换成1,就是直接代入。第三行:把第二行最后的式子中分子、分母同乘以
x+y=3xy,所以2x+2y-xy/x+y+2xy=(2×3xy-xy)/(3xy+2xy)=5/5=1
xy/(x+y)=2∴xy=2(x+y)(3xy-x-y)/(3x-5xy+3y)=[6(x+y)-x-y]/[3x+3y-10(x+y)]=5(x+y)/[-7(x+y)]=-5/7
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^
(1)∵xy+x=-1①,xy-y=-2②,∴①-②得x+y=1;(2)先把xy+x=-1,xy-y=-2的值代入代数式,得原式=-x-[2y-1+3x]+2[x+4]=-x-2y+1-3x+2x+8
设u=xy,v=y^2/x,则,uv=y^3,u^2/v=x^3,f(u,v)=(u^2/v)^(2/3)+(uv)^(2/3),所以f(y^2/x,xy)=f(v,u)=(v^2/u)^(2/3)+
答案是3/2你是不是把分母打错了教你个方法因为上下是齐次的直接令x=3y=2带入就行
3/(x-y)=1/xyx-y=3xyy-z=-3xy原式=[(y-x)-2xy]/[2(x-y)+3xy]=[(-3xy)-2xy]/[2(3xy)+3xy]=-5xy/9xy=-5/9
2(x+xy)-[(xy-3y)-x]-(-xy)=2x+2xy-xy+3y+x+xy=3x+3y+2xy=3(x+y)+2xy=3*(-2)+2*3=0
∵x-y=4xy,∴2x+3xy-2yx-2xy-y=2(x-y)+3xyx-y-2xy=8xy+3xy4xy-2xy=112.故答案为:112.
因为xy/(x+y)=1/2所以x+y=2xy原式=3(x+y)-5xy/(-x-y+3xy)=3*2xy-5xy/(-2xy+3xy)=xy/xy=1
3xy=2x+3y+5(3y-2)(x-1)=7所以3y-2=7x-1=1得y=3x=2xy=6或3y-2=1x-1=7得y=1x=6xy=6所以xy=6
∵x-y/xy=3∴x-y=3xy2x+3xy-2y/x-2xy-y=[2(x-y)+3xy]/[(x-y)-2xy]=(6xy+3xy)/(3xy-2xy)=9xy/(xy)=9
原式=-xy²(x²y^4-xy²-1)∵xy²=-2原式=2((-2)²-(-2)-1)=10