|x-y 2|-根号x y-2=0,求x²-y²

来源:学生作业帮助网 编辑:作业帮 时间:2024/07/18 16:47:47
若x2-2xy-y2-x+Y-1=0 求x-y的值

x²-2xy+y²-x+y-1=0(x-y)²-(x-y)-1=0[x-y-(1+√5)/2][x-y-(1-√5)/2]所以x-y=(1+√5)/2或x-y=(1-√5

化简(x-yx2-2xy+y2-xy+y2x2-y2)•xyy-1= ___ .

原式=[x-y(x-y)2-y(x+y)(x+y)(x-y)]•xyy-1=(1x-y-yx-y)•xyy-1=1-yx-y•xyy-1=-xyx-y.故答案是:-xyx-y.

已知x2-3xy-4y2的绝对值+2倍根号x2+4xy+4y2-1=0 求3x+6y的值

x²-3xy-4y²=0x²+4xy+4y²-1=0由(1)得(x-4y)(x+y)=0x=4yx=-y由(2)得(x+2y)²-1=0x+2y=1x

已知x+2y=0(x不等于0)求分式2xy+y2/x2-xy

x+2y=0x=-2y2xy+y2/x2-xy=-4y^2+y^2/4y^2+2y^2=-5/6

若x2+xy-2y2=0,则x

由x2+3xy+y2 x2+y2有意义,可知y与x不能同时为0.不妨设y≠0,由x2+xy-2y2=0,化为(x+2y)(x-y)=0,解得x=y,或x=-2y.把x=y代入,可得x2+3x

已知2x-3*根号(xy)-2y=0(x>0),则x2+4xy-16y2除以2x2+xy-9y2的值是多少?

已知2x-3*根号(xy)-2y=0(x>0),则x2+4xy-16y2除以2x2+xy-9y2的值是多少?2x-3*根号(xy)-2y=0(根号X-2根号Y)(2根号X+根号Y)=0根号X-2根号Y

已知X2+Y2+8X+6Y+25=0 求代数式X2++XY+4Y2分之X2-4Y2 减X+2Y分之X的值

X2+Y2+8X+6Y+25=0x²+8x+16+y²+6y+9=0(x+4)²+(y+3)²=0∴x+4=0y+3=0x=-4y=-3X2+4XY+4Y2分之

X2+Y2+8X+6Y+25=0 求代数式(x2-4y2/x2+4xy+4y2)-x/x+2y

X2+Y2+8X+6Y+25=0x^2+y^2+8x+6y+25=0x^2+8x+16+y^2+6y+9=0(X+4)^2+(y+3)^2=0x=-4y=-3(x^2-4y^2)/(x^2+4xy+4

已知x2+y2-10x-6y+34=0,求(x-y/x+y)2÷x2-y2/xy×x2+2xy+y2/x2-xy 的值.

你好x²+y²-10x-6y+34=0(x-5)²+(y-3)²=0每项都大于等于0,只有等于0时上式才成立,所以x=5y=3(x-y/x+y)²÷(

若x<y<0,则x2−2xy+y2+x2+2xy+y2=(  )

∵x<y<0,∴x-y<0,x+y<0.∴x2−2xy+y2=(x−y)2=|x-y|=y-x.x2+2xy+y2=(x+y)2=|x+y|=-x-y.∴x2−2xy+y2+x2+2xy+y2=-2x

已知2x=3y,求xy/(x2+y2)-y2/(x2-y2)的值

已知2x=3y,求xy/(x^2+y^2)-y^2/(x^2-y^2)的值2x=3y-->x=(3/2)yx^2=(9/4)y^2xy/(x^2+y^2)-y^2/(x^2-y^2)==(3/2)y*

解关于x,y的方程组{x2-y2+根号(x2+y2)=a xy=0

由xy=0,得x=0,或y=0当x=0时,代入方程1:-y^2+根号y^2=a,即y^2-|y|+a=0,解得|y|=[1±√(1-4a)]/2当y=0时,代入方程1:x^2+根号x^2=a,即x^2

若X2+Y2-2X-6Y+10=0 ,求(x2-y2)/xy的值

10拆成1+9X2-2X+1+Y2-6Y+9=0(X-1)2+(Y-3)2=0平方大于等于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-1=0,Y-3=0X=1,Y=

已知X2-2x+y2+6y+10=0,求(x2-2xy)/(xy+y2)的值

x²-2x+y²+6y+10=0,变换得(x-1)²+(y+3)²=0,∴x=1,y=-3∴(x2-2xy)/(xy+y2)=(1²-2*(-3))/

已知xy满足x2+y2-6x+2y+10=0,求立方根号x2-y2的值

条件变换:(x-3)^2+(y+1)^2=0即:y+1=0x-3=0所以:立方根号x2-y2=2

若根号x-y+y2-4y+4=0,求xy的值

根号x-y+y2-4y+4=0所以√x-y+(y-2)²=0x-y=0y-2=0所以x=y=2所以xy=2x2=4土豆团邵文潮为您答疑解难.如果本题有什么不明白可以追问,

已知x>y,且xy=1,求证:(x2+y2/x-y)≥2根号2

左边=[(x-y)^2+2xy]/(x-y)=(x-y)+2/(x-y)>=2根号二------均值不等式,其中x-y>0

x=1/(根号3-2),y=1/(根号3+2),求代数式(x2+xy+y2)/(x+y)的值

x=1/(√3-2)=-2-√3y=1/(√3+2)=2-√3x+y=-2√3xy=-1(x2+xy+y2)/(x+y)=[(x+y)²-xy]/(x+y)=11/(-2√3)=-11√3/

已知x2-y2=xy,且xy≠0,求代数式x2y-2+x-2y2的值.

∵x2-y2=xy,∴原式=x2y2+y2x2=x4+y4x2y2=(x2−y2)2+2x2y2x2y2=3x2y2x2y2=3.再问:先化简2a+1/a²-1÷a²-a/a

已知x2+4y2+x2y2-6xy+1=0,求 x4-y4/2x-y 乘 2xy-y2/xy-y2 除以(x2+y2/x

因为x²+4y²+x²y²-6xy+1=0(x²-4xy+4y²)+(x²y²-2xy+1)=0(x-2y)²