x y=6 xy=4 求x² y²= (x-y)²

xy/(2x+y)-xy/(2x-y)=3xy/(2x-y)+xy/(2x+y)=4；一、两公式对加得2xy/(2x+y)=7xy=(2x+y)*7/2二、把xy代入原公式求解就行了：(2x+y)*7

原式=3xy+6y-xy+6x=2xy+6(x+y)=2*(-2)+6*4=24-4=20

x^+2xy+y^+4=16(x+y)²=16-4=12(1)x^-2xy+y^=16(x-y)²=16(2)(1)-(2)2y*2x=-4xy=-1(3xy+6y^)+|x^-(

xy-12=4x+y≥2√(4xy)=4√(xy)xy-4√(xy)-12≥0(√(xy)-6)(√(xy)+2)≥0√(xy)≤-2,√(xy)≥6因为√(xy)≥0所以√(xy)≥6xy≥36所以

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

(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-6xy+3x-3y=-6×(-2)+3×1=15

5xy+4x+7y+6x-3xy-4xy+3y=5xy-3xy-4xy+4x+6x+7y+3y=（5-3-4）xy+（4+6）x+（7+3）y=-2xy+10x+10y=-2xy+10（x+y）因为x

原式=[4(x+y)-2xy]分之[(x+y)+xy]=[4(3xy)-2xy]分之[(3xy)+xy]=10xy分之2xy=5分之1

原式=3x-3y-6xy=3-12=-9

xy为正实数,则有2x+y>=2根号(2xy)即:xy-6>=2根号(2xy)设根号(xy)=t>0,则xy=t^2t^2-6>=2根号2tt^2-2根号2t-6>=0(t-3根号2)(t+根号2)>

3（xy+2y）-（xy-6x）=3xy+6y-xy+6x=2xy+6(x+y)=2×(-2)+6×4=-4+24=20

（4xy+12y)+[7x-(3xy+4y-x)]=4xy+12y+7x-3xy-4y+x=xy+8x+8y=xy+8(x+y)=(-2)+8*3=-2+24=22

(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-2xy+2x+3y-3xy-2y+2x-x-4y-xy=-6xy+3x-3y=-6xy+3*(x-y)当时,原式=-6*1+3*

先把代数式（5XY+4X+7Y)+(6X-3XY)-(4XY-3Y)化简（5XY+4X+7Y)+(6X-3XY)-(4XY-3Y)=5XY+4X+7Y+6X-3XY-4XY+3Y=-2XY+10X+1

（7xy+4x-7y）+(6x-3xy)-(3y+4xy)=7xy+4x-7y+6x-3xy-3y-4xy=10x-10y=10（x-y）=10*3=30

(5xy+4x+7y)+(6x-3xy)-(4xy-3y)=5xy+4x+7y+6x-3xy-4xy+3y=(5xy-3xy-4xy)+(4x+6x)+(7y+3y)=-2xy+10x+10y=-2x

xy/x+y=6∴xy=6(x+y)3x-2xy+3y/-x+3xy-y=[3x-12(x+y)+3y]/[-x+18(x+y)-y]=-9(x+y)/17(x+y)=-9/17再问：你在啊？再答：在

因为xy(y^2+y)-y^2（xy+2x）-3xy=xy^3+xy^2-xy^3-2xy^2-3xy=-xy^2-3xy根据x+y=4,x-y的绝对值=6,可以写出两个方程组x+y=4,x-y=6和

已知x+xy=-2,xy-y=-6则X+Y=42x｛-4y-3[（x+xy）²+x]-½[（y-xy）²-2y]=2x｛-4y-3[（-2）²+x]-&fr

已知xy+x+y=6则xy+x+y+1=（x+1）（y+1）=7=7×1因为x,y都是自然数（非负整数）所以x+1,y+1≥1所以只能x+1=7y+1=1即x=6,y=0或x+1=1,y+1=7即x=