若圆x^2+y^2+2x+4y-3=0 上有两个点与直线x+y+m=0 的距离为根号2 ,求实数m 的取值范围.有一个解答得-3

(x-y) (x-2y)+(x+2y) (x-3y)-2(x-3y) (x-4y)= x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2 3(x+y)+2(x-y)=36 4(x+y)-5(x-y)=2 （x+y）/2- x-y/2=6 3(x+y)=4(x-y) x+y/3+x-y/2=6 4(x+y)-2(x-y)=40 x*x+2x+y*y-4y+5=0 x= y= 2（x+y)-3(x-y)=3 ,4(x+y)+3(x-y)=15 {4x-3y-10=0 3x-2y=0,{3x-4(x-y)=2 2x-3y=1,{2(x+y)-(x-y)=3 (x+y)-2(x-y)=1 {3(x+y)-4(x-y)=4 {x+y/2 + x-y/6=1 x+y/2+x-y/3=6,4(x+y)-3(x-y)=-20 {(x+y)/2+(x-y)/3=6 4(x+y)-3(x-y)=-20 4(x+y)-3(x-y)=-20,2/x+y+3/x-y=6 3(x+y)-2(x-y)=11 4(x+y)+3(x-y)=9 4(x-y)-3(x-y)=9 5(x-y)+2(x-y)=11 3(x+y)-4(x-y)=11,2(x-y)+5(x+y)=27 先化简再求值(x-y)(x-2y)+(x-2y)(x-3y)-2(x-3y)(x-4y) x=4 y=5 已知x*x-4xy+4y*y=0 求[2x(x+y)-y(x+y)]/(4x*x-4xy+y*y)的值? {x+y=1 ,xy=-6{x(2x-3)=0,y=x²-1{(3x+4y-3)(3x+4y+3)=0,3x+2y=5{(x-y+2)(x+y)=0,x²+y²=8{(x+y)((x+y-1)=0,(x-y)(x-y-1)=0