圆C:x² 2x y² 4y-3=0

∵A=3X²Y-4XY²,B=-X²Y-2XY²A+2B+C=0∴C=-A-2B=-(3x²y-4xy²)+2(x²y+2xy&#

A+B=-2C带入,解放成.一定能消.2、同类项,所以N=2,M=3,之后的式子提取公因式：MN(M+N+1)再问：谢谢你、但我比较笨啦、能不能教我一下第一题哇？可以给你分的。

A=3x^2y-4xy^2B=-x^2y-2xy^2A+B=3x^2y-4xy^2-x^2y-2xy^2=2x^2y-6xy^2A+B+C=0C=-(A+B)=-(2x^2y-6xy^2)=6xy^2

y-x-2xy=0y-x=2xyx-y=-2xy(3x+xy-3y)/(y-xy-x)=[3(x-y)+xy]/[(y-x)-xy]=(-6xy+xy)/(2xy-xy)=-5xy/xy=-5

(-3x^y+2xy)-(4x^+xy)=-3x^y+2xy-4x^-xy=-3x^y+xy-4x^所以填上-3x^y+xy-4x^

3x^2+xy-2y^2=0推出(3x-2y)(x+y)=0推出x=-y或x=(2/3)y{(x+y)/(x-y)+4xy/(y^2-x^2)}/{(x+3y)*(x-y)}/x^2-9y^2推出：{

（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

（x+2）^2+|x+y+5|=0,x+2=0,x+y+5=0x=-2,y=-33(x^2)y-2(x^2)y+2xy-(x^2)y+4x^2-xy=4x^2+xy=4*(-2)^2+(-2)*(-3

y-x-2xy=0所以x-y=-2xyy-x=2xy所以原式=[3(x-y)+xy]\[(y-x)-xy]=[3×(-2xy)+xy]\(2xy-xy)=-5xy\xy=-5

解C=(-A-B)/2=-2x^2y+xy-2xy^2C-A=-7x^2y-3xy+xy^2

A=5x^2y-3xy^2+4xyB=7xy^2-2xy+x^2y2C=-A-B2C-2A=-3A-B=-15x^2y+9xy^2-12xy-7xy^2+2xy-x^2y=-16x^2y+2xy^2-

∵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|2x+1|+(y-4)^=0所以x=-1/2,y=4其余的该会了吧

2A=2[4x²-4xy+y²]=8x²-8xy+2y²2A+B=8x²-8xy+2y²+x²+xy-5y²=9x

①xy同非负时,2x-3√xy-2y=(2√x+√y)(√x-2√y)=0∴√x=2√y,x=4y②xy同负时,2x-3√xy-2y=[2√(-x)+√(-y)][√(-x)-2√(-y)]=0∴√(

2x＋y＋3=xy,①若x=1,则y不存在,此时5x＋4y无最值；②若x≠1,则y=(2x＋3)/(x－1),M=5x＋4y=5x＋(8x＋12)/(x－1)=(5x²＋3x＋12)/(x－

x+y=5xy（2x-3xy+2y)/(x+2xy+y）=[2（x+y）-3xy]/[(x+y)+2xy]=(2×5xy-3xy)/(5xy+2xy)=7xy/7xy=1再问：若x＋1/x＝3,求(x

2x-3√(xy)-2y=0(2√x+√y)(√x-2√y)=0因为x>0,xy≥0所以y≥02√x+√y>0所以√x-2√y=0,即x=4y(x^2+4xy-16y^2)/(x^2+xy-9y^2)

x+2=0,x+y+5=0x=-2,y=-33x^2y+{-2X^2y-[-2xy+(x^2y-4x^2)]-xy}=3x^2y-2X^2y+2xy-x^2y+4x^2-xy=4x^2+xy=4*4+

∵x-y=4xy∴原式=[2(x-y)+3xy]/[-(x-y)-2xy]=(8xy+3xy)/(-4xy-2xy)=11xy/(-6xy)=-11/6