已知x^2 y^2-xy=1,求dy dx

x³y+2x²y²+xy³=xy(x²+2xy+y²)=xy(x+y)²=1/2(2)²=2

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

分子分母同时除以xy则分子得2（1/x+1/y）-3=2*5-3=7分母得3（1/x+1/y）+2=3*5+2=17则得7/17

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

1/x+1/y=5(x+y)/x*y=5x+y=5*x*yx+2xy+y=x+y+2xy=5xy+2xy=7xy2x-3xy+2y=10xy-3xy=7xy(x+2xy+y)/(2x-3xy+2y)=

因为:1/x-1/y=3(y-x)/xy=3x-y=-3xy（1）又因为：(2x+3xy-2y)/(x-2xy-y)=[2（x-y）+3xy]/[(x-y)-2xy]代入（1）得：（-6xy+3xy）

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

答：x+y=-1,xy=-2-5（x+y)+(x-y)+x(xy+y)=-5x-5y+x-y+xy(x+1)=-4x-6y+(-2)(x+1)=-4x-6y-2x-2=-6x-6y-2=-6(x+y)

1/x-1/y=3(y-x)/xy=3y-x=3xyx-y=-3xy(2X+3XY-2Y)/(X-2XY-Y)=[2(x-y)+3xy]/[(x-y)-2xy]=(-6xy+3xy)/(-3xy-2x

1/x-1/y=3∴(y-x)/xy=3,即y-x=3xy∴x-y=-3xy∴(2x-3xy-2y)/(x-2xy-y)=[2(x-y)-3xy]/[(x-y)-2xy]=(-6xy-3xy)/(-3

1/x+1/y=(x+y)/xy=8x+y=8xy2x-3xy+2y/x+2xy+y=[2(x+y)-3xy]/[(x+y)+2xy]=(16xy-3xy)/(8xy+2xy)=13xy/10xy=1

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

-5（x+y)+(x-y)+2(xy+y)=-5x-5y+x-y+2xy+2y=-4x-4y+2xy=-4（x+y）+2xy=-4×（-1）+2×（-2）=4+（-4）=0你有问题也可以在这里向我提问

由已知通分Y-X=4XY所以所求=-2(Y-X）+xy/-（Y-X）-2XY=-7XY/-6XY=7/6

x^3y-2x^2y^2+xy^3=xy(3x^2-4xy+y^2)=xy(3x-y)(x-y)x-y=-1,xy=3y(y-1)=3

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

这道题目还是在考察韦达定理的运用用伟大定理求出xy的值再代入代数式否则是求不出来的(x+y)^2=x^2+y^2+2xy=1x^2+y^2=5(x-y)^2=5-2(-2)=9下面分两种情况讨论1x-

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

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

(-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