已知x-y=1,xy=2,求x的平方y-2xy xy的平方的值

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已知x-y=1,xy=2,求x的平方y-2xy xy的平方的值
已知x+y=2,xy=1/2,求x³y+2x²y²+xy³

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

已知x-xy=8,xy-y=-9,求x+y-2xy的值

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

已知1/x+1/y=5 求2x-3xy+2y/3x+2xy+3y

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

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,求分式(2x+3xy-2y)/(x-2xy-y)

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)

已知:x-y=1,xy=-2.求:(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)的值

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

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1/x-1/y=3,求 2X+3XY-2Y/X-2XY-Y 的值//!

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

已知1/x-1/y=3,求(2x-3xy-2y)/(x-2xy-y)

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

已知1/x+1/y=8,求2x-3xy+2y/x+2xy+y的值

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

已知x(x-1)-(x^2-y)=-3求(二分之y-x+2xy)-xy 快,

x(x-1)-(x^2-y)=-3x²-x-x²+y=-3y-x=-3(y-x+2xy)/2-xy=(y-x)/2+xy-xy=(y-x)/2=-3/2

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

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

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1/x-1/y=4,求2x+xy-2y/x-2xy-y的值

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

已知1/x-1/y=4,求(2x+xy-2y)/(x-2xy-y)

∵1/x-1/y=4∴(y-x)/xy=4∴y-x=4xy∴x-y=-4xy(2x+xy-2y)/(x-2xy-y)=[2(x-y)+xy]/[(x-y)-2xy]=(-8xy+xy)/(-4xy-2

已知x-y=-1,xy=3,求x^3y-2x^2y^2+xy^3

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

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

(-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

.已知:x-y=1,xy=-2.求:(-2xy+2 x+3y)-(3xy+2y-2x)-(x+4y+xy)的值

(-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