(x∧2+y∧2-2x+1)∧2-4(4y-4xy)(x∧2-y∧2-2x+1) 因式分解.

2个回答

  • (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