Question

因式分解(x^2+y^2-2x+1)^2-(4y-4xy)(x^2-y^2-2x+1)使用换元法

Answer 1

(x^2+y^2-2x+1)^2-(4y-4xy)(x^2-y^2-2x+1)
=[(x-1)^2+y^2]^2+4y(x-1)[(x-1)^2-y^2]
令x-1=m
原式=(m^2+y^2)^2+4ym(m^2-y^2)
=m^4+4m^3y+2m^2y^2-4my^3+y^4
=(m^4+4m^3y+4m^2y^2)-(2m^2y^2+4my^3)+y^4
=(m^2+2my)^2-2(m^2+2my)y^2+(y^2)^2
=(m^2+2my-y^2)^2
=[(x-1)^2+2(x-1)y-y^2]^2 接着要不要分解了？
继续的话
=（x^2-2x+1+2xy-2y-y^2)^2
=[(x^2+2xy+y^2-2(x+y)+1-2y^2]^2
=[(x+y)^2-2(x+y)+1-2y^2]^2
=[(x+y-1)^2-2y^2]^2
=(x+y-1+根号2 *y)^2 *(x+y-1-根号2 *y）^2

Answer 2

(1+x^2+2 x (-1+y)-y (2+y))^2