x^2 9-y^2 4=1的弦被(2,1)平分

由2(X+Y)3X+3Y=24得：2(X+Y)X+Y=8①;(X+Y/2X)XY/2Y=1得：X+Y=4②;由①、②得出Y=8(1-X),进入②知X=4/7；即Y=24/7

2/x+2/y=根号24(2y+2x)/xy=2√6x+y=xy√6x/y(x-y)-y/x(x-y)=1/(x-y)[(x/y-y/x)]=1/(x-y)[(x²-y²)/xy]

你的题目有点问题我这样做了x/{y(x-y)}-y/{x(x-y)}=（x平方-y平方）/{xy(x-y)}=(x+y)/xy2/x+2/y=2(x+y)/xy=根号24

x+y/x-y=1/2取倒数x-y/x+y=2所以x-y/x+y-2x+2y/x-y=x-y/x+y-2(x+y/x-y)=2-2×1/2=2-1=1

[2x的平方-(x+y)(x-y)][(-x-y)(y-x)+2y的平方]=[2x²-(x²-y²)][(x+y)(x-y)+2y²]=(2x²-x&

圆C1；x*x+y*y+4x+y+1=0.1圆C2：X*X+Y*Y+2X+2Y+1=0.21式减2式,得2x-y=0,即y=2x.33式代入1式,得5x^2+6x+1=0,得x=-1或x=-1/5,则

(2x-y)(2x+y)+(2x-y)(y-4x)+2y(y-3x)=4x^2-y^2+2xy-8x^2-y^2+4xy+2y^2-6xy=-4x^2=-4(-1/4)^2=-1/4

1.只要把X=-2Y带入曲线方程,求出两点坐标你先做第一个吧.后面的比较麻烦,我已经好久不用那些公式了,还要找找3.先判断（3,6）与园的关系,把3和6带入圆方程,（3,6）在圆外面,我设了直线方程,

因为X-Y=1所以原式=x*1+y*（-1）+2013=x-y+2013=1+2013=2014

1)x²-y²=(x+y)(x-y)6(x-y)=24x-y=42)2m²+4mn+2n²-6=2(m+n)²-6=2*9-6=18-6=123)81

原式=(x²-2xy+y²+x²-y²)÷x=(2x²-2xy)÷x=2x-2y=-2-1=-3

x+y=1x-y=2(x+2y)(x-2y)-(2x-y)(-y-2x)=(x+2y)(x-2y)+(2x-y)(y+2x)=x²-4y²+4x²-y²=5x&

我用x2表示的x的平方1,(x-y)2=x2-2xy+y2,(x+y)(x-y)=x2-y2,所以原式=(2x2-2xy)/2x=x-y=12,=xy+y2+x2-y2-x2=xy=-13,=x2-x

解(x-y)(x+y)-(x-2y)²+x(3x-5y)-(x-y)(x-2y)=(x²-y²)-(x²-4xy+4y²)+(3x²-5xy

这是线性规划的题目根据x-y≥-1,x+y≥1,2x-y≤1画图可以发现满足的点位于三条直线所围成的三角形内.x-y=-1与2x-y=1交点为x=2,y=3即交点为(2,3)Z=(x-2y)/(x+y

(1)x²+y²-4x+10y+29=0(x-2)²+(y-5)²=0∴x-2=0,y-2=0∴x=2.y=2(2)题目打错了吧.应该是（2x-3)²

（1）x^2+y^2-4x+10y+29=0x^2-4x+4+y^2+10y+25=0(x-2)^2+(y+5)^2=0x=2y=-5剩下的自己解下面两个题目是不是写错了,没看懂

∵解方程组x+y-2=0和2x²+y²+4x-2y-1=0得x1=(-2+√7)/3,y1=(8-√7)/3x2=(-2-√7)/3,y2=(8+√7)/3∴直线与曲线两交点的距离

对待这样的题,先画出条件里的二次曲线,然后用与y-x=0平行的直线与所给曲线相切,然后切点就是最值点