x^2+xy-2y^2+2x+7y-3

来源：百度知道编辑：UC知道时间：2024/06/07 08:28:08

因式分解
x^2+xy-2y^2+2x+7y-3

要给出过程

用待定系数法，
设x^2+xy-2y^2+2x+7y-3=(x+ay+b)(x+cy+d)
把右边乘开，令同类项的系数相等，解出a,b,c,d

a+c=1..............(1)
ac= -2.............(2)
b+d=2..............(3)
bc+ad=7............(4)
bd= -3.............(5)

结果为x^2+xy-2y^2+2x+7y-3=(x+2y-1)(x-y+3)

x(x-y)(X^2-xy-y^2)-x^2y(y-x)=多少因式分解 x^2(x+y)(x-y)-xy(x+y)(y-x) x^2y-x+xy=x(xy-1+y)能不能分解因式成x(xy-1+y)? 计算(x-y)(x^2+xy+y^2)得计算：(X+Y)(X^2-XY+Y^2) xy=2,x+y=2,求x,y. 分解因式：x^-xy-2y^-x+5y-2 (xy-1)^2+(x+y-z)(x+y-zxy)因式分解代数 x^-y^/ x^-(y-z)^ 除以 x^+2xy+y^/(x-y)^-z^ 乘 x^+xy-xz/x^-xy xy/x+y=3求2x-3xy+2y/3xy-x-y的值