已知x^2-xy-2y^2=0,且x不等于,y不等于0,求代数式

x-xy=8(1)xy-y=-9(2)则有(1)-(2):X-XY-XY+Y=X+Y-2XY=8-(-9)=17

2x-3y=0所以y=2x/3代入(x^2+xy-y^2)/(2x^2+xy-y^2)得到[x^2+x*2x/3-(2x/3)^2]/[2x^2+x*2x/3-(2x/3)^2]=(x^2+2x^2/

2x^2-3xy+y^2=(2x-y)(x-y)=0则可得2x=y或x=y①当2x=y时把2x=y代入其中得x/y+y/x=x/2x+2x/x=1/2+2=2.5②当x=y时x/y+y/x=x/x+x

3x*+xy-2y*=0（3x-2y）(x+y)=0那么x=2y/3或x=-yy/x=3/2或x/y=-1(x/y)-(y/x)-(x*+y*)/(xy)=(x/y)-(y/x)-(x/y+y/x)=

由y=3xy+x得x-y=-3xy2x+5xy-2y/x-2xy-y=2（x-y）+5xy/x-y-2xy=-6xy+5xy/-3xy-2xy1/5

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

x=-2y(2xy+y^2)/(x^2-xy)=(-4y^2+y^2)/(4y^2+2y^2)=-3y^2/(6y^2)=-1/2再问：我算出来也是-1/2，可答案说是-15/28

（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

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：{

即(x-2y)²=0x-2y=0所以x=2y所以原式=(2x²+2xy-xy-y²)/(4x²-4xy+y²)=(2x²+xy-y²

x-2y=0x=2y所以xy=2y²x²=4y²所以原式=(4y²-6y²+y²)/(8y²+2y²-3y²)

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

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

∵x-y=4xy，∴2x+3xy-2yx-2xy-y=2(x-y)+3xyx-y-2xy=8xy+3xy4xy-2xy=112．故答案为：112．

x²+y²-xy+2x-y+1=[3（x+1）²+(x-2y+1）²]/4=0,由于（x+1）²>=0且(x-2y+1）²>=0,则有x+1

无解

x²-7xy+12y²=0(x-3y)(x-4y)=0x1=3yx2=4yx=3y时原式＝9y²-3y²+y²/6y²=7/6x=4y时原式

2X^2+3xy-2y^2=0(2x-y)(x+2y)=0即：2x-y=0或,x+2y=0y=2x或x=-2y当y=2x时：x+y/x-y=3x/(-x)=-3当x=-2y时：x+y/x-y=-y/(

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

0=x^2+y^2+x^2y^2-4xy+1=x^2+y^2-2xy+x^2y^2-2xy+1=(x-y)^2+(xy-1)^2,两个平方数的和等于0,所以,x=y,xy=1,带入得(x-y)^200