{2分之x y=5分之3x4y

解1/x+1/y=5即(y+x)/xy=5∴y+x=5xy∴(2x+3xy+2y)/(x-2xy+y)=[2(x+y)+3xy]/[(x+y)-2xy]=[2×5xy+3xy]/(5xy-2xy)=(

xy/(x+y)=2解得xy=2(x+y)所以(3x-5xy+3y)/(-x+3xy-y)=(3/2*xy-5xy)/(-1/2*xy+3xy)=-7/5

1/x-1/y=3x-y=-3xy(2x+3xy-2y)/(x-2xy-y)=[2(x-y)+3xy]/[(x-y)-2xy]=(-3xy)/(-5xy)=3/5

∵(x+y)分之xy=3∴xy=3(x+y)∴(-2x+3xy-2y)分之(3x-5xy+3y)=[-2(x+y)+9(x+y)]分之[3(x+y)-15(x+y)]=[7(x+y)]分之[-12(x

(5x+3xy-5y)/(x-2xy-y)=(5/y+3-5/x)/(1/y-2-1/x)分子分母同时除以xy=[-5(1/x-1/y)+3]/[-(1/x-1/y)-2]=(-15+3)/(-3-2

3xy-[2xy-2(xy-2分之3xy)+xy]+3xy=6xy-[2xy-2xy+3xy+xy)=6xy-4xy=2xy=2×3×3分之1=2

xy/(x+y)=51/x+1/y=1/5yz/(y+z)=7/21/y+1/z=2/7zx/(z+x)=41/x+1/z=1/4(xy+yz+zx)分之xyz=1/(1/x+1/y+1/z)=280

1/x-1/y=(y-x)/xy=3y-x=3xy所以x-y=-3xy所以原式=[(x-y)-2xy]分之[2(x-y)-5xy]=[(-3xy)-2xy]分之[2(-3xy)-5xy]=(-5xy)

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

1/x+1/y=(x+y)/(xy)=5x+y=5xy(2x-3xy+2y)/(x+2xy+y)=(2*5xy-3xy)/(5xy+2xy)=(10-3)/(5+2)=1

因为X+Y分之XY=2,所以XY=2(X+Y)代入后面的分子式得：13（X+Y)/5(X+Y)=13/5

(x分之1)+(y分之1)=5（x+y)/xy=5x+y=5xy代入（x+2xy+y)分之(2x-3xy+2y)=7xy分之7xy=1

因1/x+1/y=5(x+2xy+y)/(2x-3xy+2y)上下同乘以1/(xy)得[1/y+2+1/x]/[2/y-3+2/x]整理得[(1/x+1/y)+2]/[2(1/x+1/y)-3]将已知

(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

1/x-1/y=(y-x)/xy=2y-x=2xy(3x-5xy-3y)/(-x+3xy+y)={3(x-y)-5xy}/(y-x+3xy)={(-3)*2xy-5xy}/5xy=-11/5

解(-3/4x²y-3/5xy²)(-2/3xy)²=(-3/4x²y-3/5xy²)(4/9x²y²)=-1/3x^4y

1/x+1/y=5(x+y)/x*y=5x+y=5*x*yx+2xy+y=x+y+2xy=5xy+2xy=7xy2x-3xy+2y=10xy-3xy=7xy(2x-3xy+2y)/(x+2xy+y)=

1/x-1/y=3(y-x)/xy=3y-x=3xyx-y=-3xy所以原式=[5(x-y)+3xy]/[(x-y)-2xy]=(-15xy+3xy)/(-3xy-2xy)=(-12xy)/(-5xy

(3x+5xy-3y)/(x-xy-y)=(3x-3xy-3y+8xy)/(x-xy-y)=3+8xy/(x-xy-y)=3+8/(1/y-1-1/x)=3+8/(-3)=1/3

单项式有：3分之b²,-2,共2个.多项式有：2分之xy+3,5分之ab+x,共2个.整式有：3分之b²,2分之xy+3,-2,5分之ab+x,共4个.