XY 根号1-X2-Y2

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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

设x,y为实数,且x2+xy+y2=1,求x2-xy+y2的值的范围

设x2-xy+y2=Ax2-xy+y2=A与x2+xy+y2=1相加可以得到:2(x2+y2)=1+A(1)x2-xy+y2=A与x2+xy+y2=1相减得到:2xy=1-A(2)(1)+(2)×2得

在x2+2xy-y2,-x2-y2+2xy,x2+xy+y2,4x2+1+4x中,能用完全平方公式分解因式的有(  )个

①x2+2xy-y2不符合完全平方公式的特点,不能用完全平方公式进行因式分解;②-x2-y2+2xy符合完全平方公式的特点,能用完全平方公式进行因式分解;③x2+xy+y2不符合完全平方公式的特点,不

先化简,再求值:(1)2x2-5x x2 4x,其中x=-3 (2)(3x2-xy-2y2)-2(x2 xy-2y2),

:(1)2x2-5x+x2+4x,其中x=-3=3x²-x=3x(-3)²+3=27+3=30(2)(3x2-xy-2y2)-2(x2+xy-2y2),其中x=6,y=-1=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

x2+2y2-2xy-2y+1=0

X^2+2Y^2-2XY-2Y+1=0X^2-2XY+Y^2+Y^2-2Y+1=0(X-Y)^2+(Y-1)^2=0因为两个数的平方为一个非负数,所以得:X-Y=0Y-1=0所以:X=1Y=1所以:X

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

因式分解:1-x2+4xy-4y2=______.

原式=1-(x2-4xy+4y2)=1-(x-2y)2=(1+x-2y)(1-x+2y).故答案为(1+x-2y)(1-x+2y).

x2-y2=xy,xy不等于0求x2/y2=y2/x2

是求x2/y2+y2/x2=吗x2-y2=xy则x/y-y/x=1两边平方得x^2/y^2-2+y^2/x^2=1所以x^2/y^2+y^2/x^2=3

x2+y2-xy=1,则u=x2-y2的取值范围是

(x+y)^2=1+3xy(x-y)^2=1-xyu=(x+y)(x-y)|u|=√(x+y)^2√(x-y)^2=√(1+3xy)√(1-xy)=√[-3(t-1/3)^2+2/3]≤√6/3故-√

1,1,X2-1-2xy+y2

X2-1-2xy+y2=X2-2XY+Y2-1=(X2-2XY+Y2)-1=(X-Y)2-1=(X-Y-1)(X-Y+1)(x2+4y2)2-16x2y2=(x2+4y2)2-(4XY)2=(X2+4

已知x2+xy=5,xy+y2=-1,则x2-y2=______.

∵x2+xy=5,xy+y2=-1,∴(x2+xy)-(xy+y2)=x2+xy-xy-y2=x2-y2=5-(-1)=6.故填:6

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

因式分解:x3+(y-1)x2-2xy-y2

x3+(y-1)x2-2xy-y2=x^3+x^2y-x^2-2xy-y^2=x^2(x+y)-(x+y)^2=(x+y)(x^2-x-y)

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/

X2+3xy+y2因式分解

=[x+(3-√5)/2*y][x-(3-√5)/2*y]有点牵强,但这是唯一的答案了

2x2+xy-3y2+x+4y-1因式分解

2x²+xy-3y²+x+4y-1=2×x×x+x×y-3×y×y+x+4×y-1=x×2x+x×y-y×3y+x×1+y×4-1=x×(2x+y+1)-y×(3y+4)-1再问:

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