已知两圆x+y=1,x+y-2x-2y+1=0 求（1）它们的公共弦所在直线的方程（2）公共弦所在直线被圆：（x-1）+（y-1）=25/4所截得的弦长

已知x+y=a,2x-y=-2a,求[（x/y-y/x)/(x+y)-x(1/x-1/y)]/[(x+1)/y]的值 x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2 已知x-y/x+y=3,求代数式5(x-y)/x+y-x+y/2(x-y) 已知x-y=1,求[(x+2y)^2+(2x+y)(x-4y)-3(x+y)(x-y)]除以y的值大神们帮帮忙 已知：2x+y=6,x-3y=1 求7y(x-3y)(x-3y)-2(3y-x)(3y-x) 已知:【(x²+y²)-(x-y)²+2y(x-y)】/4y=1,求4x/4x²-y²-1/2x+y 已知x-y=1,求[(x+2y)²+(2x+y)(x-4y)-3(x+y)(x-y)]÷y的值 已知x,y满足约束条件：x-y+1>=0,x+y-2>=0,x 已知x-y=3,求[（x+y)(x-y)-(x-y)^2+2y(x-y)]除以的值 已知3x =2y 求x/x+y+y/x-y-y²/x²-y² 已知f(x+y,y/x)=x^2-y^2,则f(x,y)= 已知|x|=2,|y|=3,|x+y|=-(x+y)求x-y 已知x≠y,x^2-x=y^2-y,求x+y的值. 已知x>0,Y>0,如何证x^2/y+y^2/x>=x+y 已知x=2+根号3,y=2-根号3,计算代数式(x+y/x-y-x-y/x+y)乘以(1/x^2-1/y^2) 已知x/y=1/2,求2x/(x²-2xy+y² )* (x²-y²/x+y)+2y/x-y 已知x/y=1/2,求2x/(x²-2xy+y² )* (x²-y²/x+y)+2y/x-y 已知两圆x+y-2x+10y-24=0和x+y+2x+2y-8=0.求公共弦的长度