已知xy-y²分之x² 2xy y²÷A=-搜答案-神马作文网

原式=[x-y(x-y)2-y(x+y)(x+y)(x-y)]•xyy-1=（1x-y-yx-y）•xyy-1=1-yx-y•xyy-1=-xyx-y．故答案是：-xyx-y．

将xy提取公因式变成：xy（x+y）=-3*6=-18

(x-y)/(x+y)=(x-y)(x+y)/[(x+y)^2]=(x^2-y^2)/[x^2+y^2+2xy]=2xy/[x^2+y^2+x^2-y^2]=2xy/(2x^2)=y/xx^2-y^2

答：1/x+1/y=3(x-2xy+y)/(2x+xy+2y)分子分母同除以xy得：=(1/y-2+1/x)/(2/y+1+2/x)=(3-2)/(3*2+1)=1/7

-5分之2

x/y=3x=3y(x²+2xy-3y²)/(x²-xy+y²)=(9y²+6y²-3y²)/(9y²-3y²

因为x-y=4xy所以x-2xy-y=2xy2x+3xy-2y=11xyx-2xy-y分之2x+3xy-2y=5.5

x分之1-y分之1=4(xy)分之(y-x)=4y-x=4xyx-y=-4xy(x-2xy-y)分之(2x+xy-2y)=[(x-y)-2xy]分之[2(x-y)+xy]=(-4xy-2xy)分之(-

1/X-1/Y=4Y-X=4XY(2X+XY-2Y)/(X-2XY-Y)=[2(X-Y)+XY]/[(X-Y)-2XY]=7/6

1/x+1/y=-2则(x-xy+y)/(x+xy+y)分子分母同除以xy=(1/x+1/y-1)/(1/x+1/y+1)=(-2-1)/(-2+1)=3

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

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

原式=[(x+y)2(x-y)(x+y)+-4xy(x-y)(x+y)]×(x+3y)(x-3y)(x+3y)(x-y)=x-3yx+y，由已知得（3x-2y）（x+y）=0，因为x+y≠0，所以3x

产生4种配子xyxyyyx有一种y有两种xy1y2xy1xy2y1y2产生的配子如上y1、y2是一个意思所以分为一组xyxyyy=1221

x-y分之xy=-3分之1xy分之(x-y)=-3x-y=-3xyx-2xy-y分之2x+3xy-2y=(-6xy+3xy)/(-3xy-2xy)=3/5

由已知条件可得：x-y=-4xy,（也就是在等式两边乘xy后整理,即可）,代入所求的式子里分母为-4xy+7xy,即3xy；分子为-4xy-2xy,即-6xy,化简得-2

设x+y=axy=b,则b/a=3b=3a所以-X+3XY-Y分之2X-3XY+2Y可以写成(2a-3b)/(3b-a)=-7/8

xy分之【x加y】等于2分之一xy=2(x+y)【xy加x加y】分之xy=2(x+y)/[2(x+y)+x+y]=2(x+y)/[3(x+y)]=2/3

x/3-y/4=3x+2y==>4x-3y=36x+24y==>①32x=-27y==>x=-27y/32②(4x-3y)^2=(36x+24y)^2②==>16x^2+9y^2-24xy=1296x