已知x²-y²=-1,x y=2分之一,求x-y的值等于多少

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已知x²-y²=-1,x y=2分之一,求x-y的值等于多少
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

(1)已知x+y=6,xy=-3,x^2y+xy^2=

(1)已知x+y=6,xy=-3x^2y+xy^2=xy(x+y)=-3*6=-18(2)x*[2-(1/x)]+(x/x^2-2x)*(x^2-4)=x*(2x-1)/x+[x/x(x-2)]*(x

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x+y=1,xy=1/5,则x^2y=xy^2

x^2y+xy^2=xy(x+y)=1/5

已知1/x=1/y=6,则3x-2xy+3y/3xy+x+y=?

/>∵1/x=1/y=6∴x=y=1/6∴(3x-2xy+3y)/(3xy+x+y)=(3/6-2*1/6*1/6+3/6)/(3*1/6*1/6+1/6+1/6)上下同乘36=(18-2+18)/(

已知:xy+x=-1,xy-y=-2.

(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

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x+y=0,x+13y=1,求x²+12xy+13y²的值.

解题思路::∵x+y=0,x+13y=1,解得x=1/12,y=-1/12∴x²+12xy+13y²=1/144-1/12+13/144=14/144-1/12=2/144=1/72解题过程:已知x+

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=4xy,则2x+3xy-2yx-2xy-y

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

已知实数xy满足x²﹢y²-xy+2x-y+1=0求xy

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

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xy>0,证明xy+xy/1+x/y+y/x>=4

xy+1/xy>=2√(xy*1/xy)=2(当xy=1/xy即xy=1时取等号)x/y+y/x>=2√(x/y*y/x)=2(当x/y=y/x即x=y取等号)当x=y=1时可以同时满足两项的等号要求

已知xy/x+y=1/2,则代数式3x-5xy+3y/-x+3xy-y=

因为xy/(x+y)=1/2所以x+y=2xy原式=3(x+y)-5xy/(-x-y+3xy)=3*2xy-5xy/(-2xy+3xy)=xy/xy=1

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