用待定系数法解题:2x^2+3xy+y^2-4x-3y+2
待定系数法因式分解:(1)x方+2xy+y方+x+y-2 (2)x方+xy-2y方+2x+7y-3
用待定系数法分解因式1.多项式3x2+5xy-2y2+x+9y+n能被3x-y+4整除,则n=2.2次3项式当x=1时其
已知1/X+2/X=2,求代数式(4X+3XY+2Y)/(-4X+8XY-2Y)的值.写出解题思路和过程..
(x-2xy)*(-xy+2y*y)-(3x*x-2xy)(x-9xy+6y*y)
(-3x^y+2xy)-( )=4x^+xy
(x+y)(x-y)-(2xy+3y)平方
[x(x^2y^2-xy)-y(x^2-x^3y)]/3x^2y
已知3x^2+xy-2y^2=0,求{(x+y)/(x-y)+4xy/(y^2-x^2)}/{(x+3y)*(x-y)}
[-(x-y)^2/xy]^-4*(y^2-xy/x)^3*(x^4/y^10)÷(xy-y^2/x)^-5=-y(y-
1、(-7x^y)(2x^y-3xy^3+xy) 2、((x-y)^6)/((y-x)^3)/(x-y)
化简:x-y/x+3y÷x^-y^2/x^2+6xy+9y^2-(xy/x+y)
x^3y(-4y)^2+(-7xy)^2*(-xy)-5xy^3*(-3x)^2