圆C:x^2+y^2-2x-2y-7=0,设P是该圆的过点（3,3）的弦中点,则动点P的轨迹方程是?

x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2 如果A={（x,y）|x,y+2>0},B={（x,y）|2x+3y-6>0},C={（x,y）|x-4 a(x-y)(x+y)-b(y-x)(x+y)-c(x-y)^2(y+x)提取公因式 7y(x-3y)-2(3y-x)=0 .x=?,y=? 曲线C：2x^+y^-x+2y-3=0关于y轴对称 x+2y=2x+y+1=7x-y 求：2x-y? 已知x>0,Y>0,如何证x^2/y+y^2/x>=x+y 解方程组(x+y)(x-2y)=-5 (x-y)(x+2y)=7 x^2+y^2=15,x+y=7,x＞y,求x-y (X,Y) f(x,y)={12y^2,0 x*x+2x+y*y-4y+5=0 x= y= {6(x-y)-7(x+y)=21 {2(x-y)-5(x+y)=-1 当x+y=2时7(x+y)一(x+y)+(x+y)=? 是否存在常数c,使得不等式x/(2x+y)+y/(x+2y)≤c≤x/(x+2y)+y/(2x+y)对任意正数x,y恒成立[x/(2x+y) +y/(x+2y)]-[x/(x+2y) +y/(2x+y)] =[x(x+2y)+y(2x+y)-x(2x+y)-y(x+2y)]/[(x+2y)(2x+y)] =[x^2+2xy+2xy+y^2-2x^2-xy-xy-2y^2]/[(x+2y)(2x+y)] =(2xy 是否存在常数c,使得不等式x/(2x+y)+y/(x+2y)≤c≤x/(x+2y)+y/(2x+y)对任意正数x,y恒成立[x/(2x+y) +y/(x+2y)]-[x/(x+2y) +y/(2x+y)] =[x(x+2y)+y(2x+y)-x(2x+y)-y(x+2y)]/[(x+2y)(2x+y)] =[x^2+2xy+2xy+y^2-2x^2-xy-xy-2y^2]/[(x+2y)(2x+y)] =(2xy 设Z=X+Y,其中X,Y满足X+2Y>=0,X-Y 若(x-y)/(x+y)=3 则(x+2y)/(6x-7y)等于 化简y(x+y)+(x+y)(x-y)-x^2