如果方程组y2=2x y=x b

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如果方程组y2=2x y=x b
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.

解方程组 ﹛x2+y2=25和xy=-12

还有一种情况没照出来x=-4y=3

如果M=3x2-2xy-4y2,N=4x2+5xy-y2,则8x2-13xy-15y2等于(  )

A、原式=-6x2-19xy-5y2;B、原式=2x2-9xy-7y2;C、原式=x2-16xy-10y2;D、原式=8x2-13xy-15y2.故选D.

已知向量ab不共线,实数xy满足向量等式3xa+(10-y)b=2xb(4y+7)a+2xb

移向有3xa+(10-y-2x)b-2x(4y+7)ab=0因为不共线所以x=0y=10

解方程组(1):xy-y-2x+2=0 (2):x2-2x+y2-4x=-5

化简(1)得y(x-1)=2(x-1)然后分类讨论(因为要看x-1能不能直接两边同除)当x=1时,根据(2)式算出y=0;当x不等于1时y(x-1)=2(x-1)就可以两边同除以(x-1)得y=2,代

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

已知x2+xy=2,y2+xy=5,则12x2+xy+12y2=___.

∵x2+xy=2,y2+xy=5,∴x2+2xy+y2=7,则原式=12(x2+2xy+y2)=72,故答案为:72

已知x2+xy=3,xy+y2=-2,则2x2-xy-3y2=______.

有x2+xy=3可得,2x2+2xy=6  (1),有xy+y2=-2得,3xy+3y2=-6 (2),根据分析,(1)-(2)可得,2x2-xy-3y2=6-(-6)=

X2+Y2+XY=2,求X2+Y2-XY取值范围

x^2+xy+y^2=2≥3xyxy≤2/3-2xy≥-4/3,x^2-xy+y^2=x^2+xy+y^2-2xy=2-2xy≥2/3当且仅当x=y时取等号再问:>=2/3且

解方程组 x2+y2=10 x+xy+y=7

∵x^2+y^2=10,x+xy+y=7∴7-xy=x+y,且2xy≤x^2+y^2=10,∴(7-xy)^2=(x+y)^2=x^2+y^2+2xy=10+2xy,且xy≤5∴49-14xy+(xy

解方程组x2+y2=10,xy=3

答:x²+y²=10xy=3,y=3/x代入上式得:x²+(3/x)²=10整理得:(x²)²-10*x²+9=0(x²

解方程组 x2+y2=4 xy-y2+4=0

第一个式子:x的平方等于4-y的平方第二个式子:4-y的平方等于-xy说明x=-y所以x=根号2;y=-根号2或则x=-根号2;y=根号2

二元二次方程组x2+y2+2xy=9 (x-y)2-3x+3y+2=0

x2+y2+2xy=9(x+y)²=9x+y=3或-3(x-y)2-3x+3y+2=0(x-y)²-3(x-y)+2=0(x-y-1)(x-y-2)=0x-y=1或2所以组成4个方

解方程组x2+y2=20,2x2-3xy-2y2=o

由2x²-3xy-2y²=0得2-3(y/x)-2(y/x)²=0(2+y/x)*(1-2y/x)=0得y/x=1/2或-2即y=1/2x或y=-2x代入x²+

解方程组x2+xy=12 xy+y2=4

x^2+xy=12xy+y^2=4因式分解下,得x(x+y)=12.y(x+y)=4两个方程相加,得(x+y)^2=16所以x+y=±4当x+y=4时,代入x(x+y)=12.y(x+y)=4解得x=

解二元二次方程组①x2-y2=3②x2+y2+2xy+x+y=12方程组如上:求详解

因为X^2-Y^2=(X+Y)(X-Y)x^2+y^2+2xy=(X+Y)^2这个题目可以因式分解成1,(X+Y)(X-Y)=32,(X+Y)^2+(X+Y)=12设X+Y=AX-Y=B那么方程变成A

如果x2+xy=2,xy+y2=-1,则x2-y2=______,x2+2xy+y2=______.

(1)∵x2-y2=x2+xy-xy-y2=x2+xy-(xy+y2)而x2+xy=2,xy+y2=-1,∴x2-y2=2-(-1)=3;(2)∵x2+2xy+y2=x2+xy+xy+y2,而x2+x

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