x y=5,x-y=1,求x的平方加y的平方

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1/x+1/y=5,求2x-3xy+2y/x²+xy+y的值

1/x+1/y=5(y+x)/xy=5x+y=5xy2x-3xy+2y/x+xy+y=2(x+y)-3xy/(x+y)+xy=10xy-3xy/5xy+xy=7xy/6xy=7/6如果本题有什么不明白

已知x-y=2xy,求代数式3x-5xy-3y/x+xy-y的值

x-y=2xy3x-5xy-3y/x+xy-y=[3(x-y)-5xy]/[(x-y)+xy]=(6xy-5xy)/(2xy+xy)=1/31/x+1/y=4y+x=4xyx-5xy+y/2x+xy+

已知,xy/x+y=3,求代数式3x-5xy+3y/-x+3xy-y的值

xy/x+y=3xy=3(x+y)3x-5xy+3y/-x+3xy-y=(3x+3y)-5*3(x+y)/[-x-y+3*3(x+y)]=-12(x+y)/8(x+y)=-3/2望采纳,谢谢!

已知1/x+1/y=5,求x+2xy+y/2x-3xy+2y的值

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)=

已知1/x-1/y=3,求x-2xy-y/5x+3xy-5y的值

1/x-1/y=3(Y-X)/XY=3Y-X=3XYX-Y=-3XY(x-2xy-y)/(5x+3xy-5y)=(-3XY-2XY)/(-15XY+3XY)=-5XY/-12XY=5/12

已知x+y=-1,xy=-2,求代数式-5(x+y)+(x-y)+x(xy+y)的值

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)

已知x+y=3,xy=1,求代数式5xy+4x+7y+6x-3xy-4xy+3y的值.

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

若1/x+1/y=5,求3x+4xy+3y/x+3xy+y的值

1/x+1/y=5两边乘以xy得y+x=5xy∴x+y=5xy3x+4xy+3y/x+3xy+y=[3(x+y)+4xy]/[(x+y)+3xy]=(15xy+4xy)/(5xy+3xy)=19xy/

已知x+y=-1,xy=-2,求代数式-5(x+y)+(x-y)+2(xy+y)的值

-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

已知xy=2(x+y),求(5x-xy+5y)/(3xy-x-y)的值

(5x-xy+5y)/(3xy-x-y)=[5(x+y)-xy]/[3xy-(x+y)]=[5(x+y)-2(x+y)]/[6(x+y)-(x+y)]=1/2

已知1/x-1/y=5,求分式y+xy-x/2x+7xy-2y的值.

解题方法:分子、分母同时除一个xy(y+xy-x)/(2x+7xy-2y)=[(y+xy-x)/xy]/[(2x+7xy-2y)/xy]=(1/x+1-1/y)/(2/y+7-2/x)=[(1/x-1

已知1/x-1/x=3,求x-xy-y分之5x+xy-5y的值

设1/x-1/y=t(5x+xy-5y)/(x-xy-y)=(1-5(1/x-1/y))/(-1-(1/x-1/y))=(5t-1)/(t+1)至于是1/x-1/y=3还是1/y-1/x=3都差不多

已知1/X-1/y=5,求分式(y+xy-x)/(2x+7xy-2y)的值

(y+xy-x)/(2x+7xy-2y)(分子、分母同除一个xy)=(1/x+1-1/y)/(2/y+7-2/x)=6/(-10+7)=-2

已知X+Y=-1,xy=-2,求代数式-5(x+y)+(x-y)+(xy+y)的值

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-

已知1/X-1/Y=-2,求3X-5XY-3Y/-X+3XY+Y的值

分子,分母同时除xy(3/y-5-3/x)/(-1/y+3+1/x)=(6-5)/(3-2)=1

已知1/x-1/y=3,求代数式5x+xy-5y除以x-xy-y的值.

(5x+xy-5y)/(x-xy-y)=(5/y+1-5/x)/(1/y-1-1/x)=(1-3)/(-1-3)=1/2

5x的平方+4xy+y的平方+2x+1=0 求xy

5x²+5xy+y²+2x+1=04x²+y²+4xy+2x+x²+1=0(2x+y)²+(x+1)²=0得{2x+y=0---①

若1/x-1/y=3,求5x+2xy-5y/x-4xy-y的值

得y-x=3xyx-y=-3xy原式=-15xy+2xy/-3xy-4xy=-13/-7=13/7

已知1/x+1/y=5,求(3x-5xy+3y)/(x+xy+y)的值?

1/x+1/y=5(x+y)/xy=5x+y=5xy(3x-5xy+3y)/(x+xy+y)=(15xy-5xy)/(5xy+xy)=10/6=5/3

已知:1/X+1/Y=5,求 2x-3xy+2y/x+2xy+y的值,

因为1/x+1/y=5y/xy+x/xy=5(x+y)/xy=5所以x+y=5xy所以(2x-3xy+2y)/(x+2xy+y)=[2(x+y)-3xy]/[(x+y)+2xy]=(2×5xy-3xy