(2x−y 2)(y−2x 2)=?

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已知实数x,y满足x2+y2-2x+4y=0,则x2+y2的最小值是

再问:对不起题目打错了,是已知实数x,y满足x2+y2-2x+4y-20=0,则x2+y2的最小值是A.30-10√5B.5-5√5C.5D.25再答:更改后的答案:

已知x2+y2-2x-4y+5=0,分式yx−xy

∵x2+y2-2x-4y+5=0,∴x2-2x+1+y2-4y+4=0,(x-1)2+(y-2)2=0,∴x=1,y=2,∴yx−xy=2-12=1.5;故答案为:1.5.

已知圆x2+y2-2x-2y+1=0求x2+y2的最大值

(x-1)^2+(y-1)^2=1令x-1=sinay-1=cosa则x=1+sina,y=1+cosax^2+y^2=1+2sina+(sina)^2+1+2cosa+(cosa)^2=3+2(si

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

已知两圆x2+y2-10x-10y=0,x2+y2+6x-2y-40=0,

(1)x2+y2-10x-10y=0,①;x2+y2+6x-2y-40=0②;②-①得:2x+y-5=0为公共弦所在直线的方程;(2)弦心距为:|10+5−5|22+12=20,弦长的一半为50−20

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*

解方程组:y2=x3-3x2+2x;x2=y3-3y2+2y

y^2=x^3-3x^2+2xx^2=y^3-3y^2+2y两式相减得:y^2-x^2=(x^3-y^3)-3(x^2-y^2)+2(x-y)(x-y)(x^2+xy+y^2-2x-2y+2)=0所以

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=

求函数f(x,y)=(x2+y2)2-2(x2-y2)的极值

x=0或x=整负根号下1-y方

已知2x2-xy-3y2=0,求x−yx+y

2x2-xy-3y2=0,(2x-3y)(x+y)=0,解得:2x-3y=0或x+y=0(分母为0,舍去),解得:x=3y2,则x−yx+y=3y2−y3y2+y=y5y=15.

已知x,y为实数,且(x2 +y2)(x2 +y2+2)=3.求x2 +y2的值

设t=x2+y2(t大于等于0)则t(t+2)-3=0(t+3)(t-1)=0t=-3(舍去)或t=1所以,x2+y2=1

已知x2+y2+4x-6y+13=0 求x2-2x\x2+3y2

即(x²+4x+4)+(y²-6y+9)=0(x+2)²+(y-3)²=0所以x+2=y-3=0x=-2,y=3所以原式=(4+4)/(4+27)=8/31

已知实数x.y满足(x2+y2)(x2+y2-1)=2,求x2+y2的值

可设x²+y²=t.则t(t-1)=2.===>t²-t-2=0.===>(t-2)(t+1)=0.===>t=2.即x²+y²=2.

已知:x=3,y=-5,x2-2xy+y2/x2-y2的值是

x²-2xy+y²/x²-y²=(x-y)²/(x-y)(x+y)=(x-y)/(x+y)因为x=3,y=-5,所以(3-(-5))/(3+(-5))

已知方程x2 +y2+4x-2y-4=0,求x2 +y2的最大值

原式可化简为(x+2)^2+(y-1)^2=9这是一个以(-2,1)为半径的圆所以x^2+y^2的最大值就是圆上一点到原点的最大距离就是圆心到原点的距离加上半径等于3+根号5

已知实数x,y满足x2+y2-2x+2y=6求x2+y2的最值

x2+y2-2x+2y=6(x-1)²+(y+1)²=2²所以可设x=1+2cosay=-1+2sina于是x²+y²=(1+2cosa)²

已知x2+y2=20,x2+xy-2y2=2(x+2y),求x+y的值

因为x2+xy-2y2=2(x+2y)即(x-y)(x+2y)=2(x+2y)所以x-y=2(x-y)2=x2-2xy+y2=4所以xy=(x2+y2-4)/2=(20-4)/2=8(x+y)2=x2

已知x2+4x+y2-2y+5=0,则x2+y2=______.

x2+4x+y2-2y+5=0,x2+4x+4+y2-2y+1=0,(x+2)2+(y-1)2=0,x+2=0,y-1=0,解得x=-2,y=1,x2+y2=5,故答案为:5.