已知x=-1,y=7/2或x=2 y=2 都是方程ax+by=6的解 求a,b

已知：2x+y=6,x-3y=1 求7y(x-3y)(x-3y)-2(3y-x)(3y-x) 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 已知x-y/x+y=3,求代数式5(x-y)/x+y-x+y/2(x-y) x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2 已知X和Y满足3X+4Y=2,X-Y-1/7 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:【(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=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x,y满足约束条件：x-y+1>=0,x+y-2>=0,x 已知|x|=2,|y|=3,|x+y|=-(x+y)求x-y 已知f(x+y,y/x)=x^2-y^2,则f(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 若x-y=2(x+y),求代数式x-y/x+y+x+y/x-y求代数式 x+y分之x-y+x-y分之x+y 已知:y=ax的3次方+bx+3,当x=-1/2时,y=-7,试求x=1/2时,y的值 把下列各式分解因式 x(x+y)(x-y)-x(y+x)² 2 已知2x+y=6,x-3y=1,求7y（x-3y）²把下列各式分解因式x(x+y)(x-y)-x(y+x)²7、已知2x+y=6,x-3y=1,求7y（x-3y）²-2（3y-x）³的值 已知方程组{(X+Y)-3(X-Y)=7,2(X+Y)+(X-Y）=-7}则(X+Y)(X-Y)=