从圆C：x^2+y^2-2x-2y-2=0外引一点P(x0,y0)向圆引切线,切点为M,O为坐标原点,且有| PM|=|PO|,求PM最小时从圆C：x^2+y^2-2x-2y-2=0外引一点P(x0,y0)向圆引切线,切点为M,O为坐标原点,且有| PM|=|PO|,求|PM|最小的P点

x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2 从M={(x,y)||x-2|+|y-2| a(x-y)(x+y)-b(y-x)(x+y)-c(x-y)^2(y+x)提取公因式 化简y(x+y)+(x+y)(x-y)-x^2 已知等式3x-2y=3y-2x-1,则x与y的大小关系是?A.x大于y B.x小于y C.x等于y D.不能确定 从这几个选项中选 在平面直角坐标系中,集合C＝{(x,y)/(y=x)},表示直线y=x,从这个角度看.集合D＝『（x,y)/{2x-y=1,x+4y=5}, 1.在平面直角坐标系中,集合C=｛(x,y)|y=x｝表示直线y=x,从这个角度看,集合D=｛（x,y）|｛2x-y=1 x+4y=51.在平面直角坐标系中,集合C=｛(x,y)|y=x｝表示直线y=x,从这个角度看,集合D=｛（x,y）|｛2x-y=1 x+4y=5 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 (x+y)(x-y)-(x-y)^2如题 (x+y)(x-y)-(x-y)^2 圆C：x^2+y^2+2x-4y+3=0,若不过原点的直线l与圆C相切,在x,y轴上截距相等,求直线l.从圆C外一点P(x,y)向...圆C：x^2+y^2+2x-4y+3=0,若不过原点的直线l与圆C相切,在x,y轴上截距相等,求直线l.从圆C外一点P(x,y) 【2(x-y)(x-y)(x-y)-8(x-y)(x-y)(x+y)+6y(x-y)(x-y)]/2(x-y)(x-y) 因式分解x(x+y)(x-y)-x(x+y)^2 填空：x-y/x+y=( )/x^2-y^2 解微分方程y`=1/(x-y)+1答案(x-y)^2+2x=c （x-y)2•(y-x)³(x-y)=? C语言中printf((%d%d,x,y,(x,y))x=1,y=2 y+y'-2y=2x