过圆x² y²-x y-2=0圆x²+y²=5的交点
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![过圆x² y²-x y-2=0圆x²+y²=5的交点](/uploads/image/f/7455779-35-9.jpg?t=%E8%BF%87%E5%9C%86x%C2%B2+y%C2%B2-x+y-2%EF%BC%9D0%E5%9C%86x%C2%B2%2By%C2%B2%EF%BC%9D5%E7%9A%84%E4%BA%A4%E7%82%B9)
3x*+xy-2y*=0(3x-2y)(x+y)=0那么x=2y/3或x=-yy/x=3/2或x/y=-1(x/y)-(y/x)-(x*+y*)/(xy)=(x/y)-(y/x)-(x/y+y/x)=
对.前提是x不等于y
y-x-2xy=0y-x=2xyx-y=-2xy(3x+xy-3y)/(y-xy-x)=[3(x-y)+xy]/[(y-x)-xy]=(-6xy+xy)/(2xy-xy)=-5xy/xy=-5
(x+y-2xy)(x+y-2)+(1-xy)^2=(x+y)^2-2(1+xy)(x+y)+4xy+1-2xy+x^2y^2=(x+y)^2-2(1+xy)(x+y)+(1+xy)^2=(x+y)^
(3xy)平方/(-xy)+(x-2y)平方-(x+2y)(x-2y)=-9xy+(x-2y)(x-2y-x-2y)=-9xy+(x-2y)(-4y)=-9xy-4xy+8y²=8y
=xy(x-y)×[-xy/(x-y)]=-(xy)²=-x²y²
即y=-2x所以原式=(x²-2x²+4x²)/(-4x²-x²)=3x²/(-5x²)=-3/5
由题设可知y=xy-1,∴x=yx3y=x4y-1,∴4y-1=1,故y=12,∴12x=x,解得x=4,于是x+y=4+12=92.故答案为:92.
【x²+xy/(x-y)】/【xy/(x-y)】=【x²(x-y)/(x-y)+xy/(x-y)】/【xy/(x-y)】={【x²(x-y)+xy】/(x-y)}/【xy
(x*x+y*y)(x*x+y*y)-4x*x*y*y=(x^4-2x^2y^2+y^4)=(x^2-y^2)^2=0x^2=y^2x/y=±1(x*x+5xy+y*y)/(x*x+2xy+y*y)=
3x^2+xy-2y^2=0推出(3x-2y)(x+y)=0推出x=-y或x=(2/3)y{(x+y)/(x-y)+4xy/(y^2-x^2)}/{(x+3y)*(x-y)}/x^2-9y^2推出:{
即(x-2y)²=0x-2y=0所以x=2y所以原式=(2x²+2xy-xy-y²)/(4x²-4xy+y²)=(2x²+xy-y²
y-x-2xy=0所以x-y=-2xyy-x=2xy所以原式=[3(x-y)+xy]\[(y-x)-xy]=[3×(-2xy)+xy]\(2xy-xy)=-5xy\xy=-5
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+
(x+y)(x-y)-x(y+1)-y(y+1)=(x+y)(x-y)-(x+y)(y+1)=(x+y)(x-2y-1)
lg(xy)=0xy=1则:|x|=1而Y=X,|x|不等于Y所以x=-1,y=-1原式=log8(2)=1/3
(x-y)/x²÷(x²-2xy+y²)/xy×(xy-x²)=(x-y)/x²÷(x+y)²/xy×x(y-x)=(x-y)²/
x+y=5xy(2x-3xy+2y)/(x+2xy+y)=[2(x+y)-3xy]/[(x+y)+2xy]=(2×5xy-3xy)/(5xy+2xy)=7xy/7xy=1再问:若x+1/x=3,求(x
原式=-x²y+2xy²+2x²y²-4xy³-3x³+27x³y-18x²y²+2x²y-18x&
利用公式法逆向因式分解(x+y)(x^2-xy+y^2)=x^3+y^3(x-y)(x^2+xy+y^3)=x^3-y^3都是公式可以自己乘开来试下可以抵消的(x^2-xy+y^2)和(x^2+xy+