(1+y)^2-2x^2(1+y^2)+x^4(1-y)^2 (可用添拆项法) 分解因式

来源：学生作业帮助网编辑：六六作业网时间：2024/06/01 09:34:44

(1+y)^2-2x^2(1+y^2)+x^4(1-y)^2(可用添拆项法)分解因式(1+y)^2-2x^2(1+y^2)+x^4(1-y)^2(可用添拆项法)分解因式(1+y)^2-2x^2(1+y

(1+y)^2-2x^2(1+y^2)+x^4(1-y)^2 (可用添拆项法) 分解因式
(1+y)^2-2x^2(1+y^2)+x^4(1-y)^2 (可用添拆项法) 分解因式

(1+y)^2-2x^2(1+y^2)+x^4(1-y)^2 (可用添拆项法) 分解因式
原式=[(1+y)-x^2(1-y)]^2-4*x^2*y^2
=(1+y-x^2+x^2*y)^2-(2xy)^2
=(1+y+2xy-x^2+x^2*y)(1+y-2xy-x^2+x^2*y)
=[(1-x^2)+(yx+y)+(x^2*y+xy)][(1-x^2)-(xy-y)+(x^2*y-xy)]
=(x+1)(x-1)(x-xy+1+y)(x-xy-1-y)

[(-x-y)(-x+y)-(x+y)^2-x(y-y^2)}÷1/2y 1、y-x/x²-y²2、(x-y)(y-z)(z-x)/(z-y)(y-x)(x-z) (x-y)(2x-y-1) [(-2x-y)^2-(2x+y)(2x-y)]÷2y-1/2y 2y(x+二分之一y)-[(x+y)(x-y)+2y(y+x)],其中|x-1|=2 （x-y)(x+y)-(x+y)^2+2y(y-x),其中x=1,y=3. 求(1+y)-2x(1+y)+x(1-y) 把4（x-y+1)+y(y-2x)因式分解 y(y+2)+4x(x-y-1)因式分解是多少 (x-y/1-x+y/1)÷2y/y-x计算错了，是(1/x-y-1/x+y)÷2y/y-x x²/xy-x/y 3x/4x+y - x-2y/4x+y (1-y/y+x)÷x/y²-x² (x/x+y-x/y-x-2xy/x.x-y.y)/(1/x-1/y) 分式化简对于(y一撇）y`=y/x(y-1)求导,我求的结果是y``=-y^2-y`x+y/x^2(y-1)^2 怎么不对啊,我的求导过程是：y`x(y-1) - [x(y-1)]` y y`xy-y`x-(y-1+xy`)y -y^2-y`x+yy``=-------------------------=----------------------=-----------------x^2 (y-1 已知x+y=a,2x-y=-2a,求[（x/y-y/x)/(x+y)-x(1/x-1/y)]/[(x+1)/y]的值若2/x-1/y=3,求[y/x-y/x-y(x-y/x-x+y)]/x-2y/x的值 x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2 化简(2x+y)(2x-y)-(x+2y)(x-2y-1) x+2y=2x+y+1=7x-y 求：2x-y?