数学因式分解

显示全部楼层 · 2010-8-16 11:36:11

（x+y）^3+2xy（1-x-y）-1
过程要详细点，先谢谢了

千问 · 2010-8-16 11:36:11

设：u=x+y,v=xy,用换元法分解因式.解：=u^3+2v（1-u）-1=(u^3-1)-2v（u-1）=（u-1）（u^2+u+1)-2v(u-1)=(u-1)(u^2+u+1-2v)再带回去=(x+y-1)(x^2+y^2+x+y+1) 或者：（x+y)^3+2xy(1-x-y)-1=(x+y)^3-1+2xy(1-x-y)=(x+y-1)[(x+y)^2+(x+y)+1)-2xy(x+y-1)=(x+y-1)[(x+y)^2+x+y+1-2xy]=(x+y-1)(x^2+2xy+y^2+x+y+1-2xy)=(x+y-1)(x^2+y^2+x+y+1）

千问 · 2010-8-16 11:36:11

(x+y)^3+2xy（1-x-y）-1=[(x+y)^3-1]-2xy(x+y-1)=(x+y-1)[(x+y)^2+(x+y)+1]-2xy(x+y-1)=(x+y-1)[(x+y)^2+(x+y)+1-2xy]=(x+y-1)(x^2+y^2+x+y+1)

千问 · 2010-8-16 11:36:11

(x+y)^3-1+2xy(1-x-y)=(x+y-1)[(x+y)^2+(x+y)+1]-2xy(x+y-1)=(x+y-1)（x^2+y^2+x+y+1）