已知x y÷xy=1÷x-y,求x² y²÷xy的值.

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

令F（XY）=1/XY+XY,当XY=1的时候,F（XY)=2,最小.（可由函数图形象得出）.XY趋于正无穷大的时候F(XY)趋于正无穷大,XY无限趋于零的时候F(XY)趋于正无穷大.所以XY越接近1

=-x-(2y-2+3x)+2(x+4)=-x-2y+2-3x+2x+8=-4x-2y+10

绝对值大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立所以两个都等于0所以x+y+1=0xy+3=0xy=-3x+y=-1两边平方x^2+2xy+y^2=(-1)^2x^2+y^2=1-

-xy（x^2y^5-xy^3-y)=-(xy^2)^3+(xy^2)^2+xy^2=-(-2)^3+4-2=8+4-2=10

x^2-xy-2y^2=x^2-y^2-xy-y^2=(x-y)(x+y)-y(x+y)=(x+y)(x-2y)=0x+y=0或x-2y=0x=-y或x=2yy÷x=-1或y÷x=1/2

(-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

解(x+1)平方+/y-1/=0∴x+1=0,y-1=0∴x=-1,y=1∴2(xy-5xy平方)-(3xy平方-xy)=(2xy+xy)+(-10xy平方-3xy平方)=3xy-13xy平方=3×(

（x+y+2xy+2y）÷（2xy+y）=(x+y/2xy)+(2xy/2xy)+(2y/y)=x+y/2xy+3已知1/x+1/y=2∴x+y/2xy=1/2(1/X+1/y)=1/2×2=1则原式

3/(x-y)=1/xyx-y=3xyy-z=-3xy原式=[(y-x)-2xy]/[2(x-y)+3xy]=[(-3xy)-2xy]/[2(3xy)+3xy]=-5xy/9xy=-5/9

（x+y)（x-y)-y^2+（x-y)^2-（6x^2y-2xy^2)÷2y=（x+y)（x-y)-y^2+（x-y)^2-2y（3x^2-xy)÷2y=（x+y)（x-y)-y^2+（x-y)^2

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

3-y=0x+y=0∴x=-3y=3[2(x+y)-3(xy+4)]÷1／xy=[2×0-3(-9+4)]÷(-1/9)=15×（-9）=-135

1/x+1/y=（x+y)/xy=5x+y=5xy所以原式=[2(x+y)-3xy]÷[(x+y)+2xy]=[2(5xy)-3xy]÷[(5xy)+2xy]=7xy÷7xy=1再问：1/x+1/y=

=xy(1-2y^2)=-0.119616

(x^4-y^4)÷(x^2+y^2/xy)÷(6x-6y)=(x²+y²)(x+y)(x-y)×[xy/(x²+y²)]×1/6(x-y)=xy(x+y)/6

(X²+3XY+2Y²)÷(X²Y+2XY²)=(x+2y)(x+y)÷[xy(x+2y)]=(x+y)/(xy)=12/9=4/3

无论是在实数范围内还是在复数范围内研究,这个问题都是有缺陷的.首先可以通过实例证明这个题目没有确定实数解取x=y=1得结果为1/2取x=1/2y=2得结果为22/29我估计这个题目会不会是把条件写错了

(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-2xy+2x+3y-3xy-2y+2x-x-4y-xy=-6xy+3x-3y=-6*(-2)+3*1=15不懂可追问,有帮助请采