先阅读下列因式分解的过程,再回答所提出的问题:

1个回答

  • (1)1+ax+ax(1+ax)+ax(1+ax) 2+…+ax(1+ax) n

    =(1+ax)(1+ax)+ax(1+ax) 2+…+ax(1+ax) n

    =(1+ax) 2+ax(1+ax) 2+…+ax(1+ax) n

    =(1+ax) 2(1+ax)+…+ax(1+ax) n

    =(1+ax) 3+…+ax(1+ax) n

    =(1+ax) n(1+ax)

    =(1+ax) n+1

    (2)x-1-x(x-1)+x(x-1) 2-x(x-1) 3+…-x(x-1) 2003+x(x-1) 2004

    =(x-1)(1-x)+x(x-1) 2-x(x-1) 3+…-x(x-1) 2003+x(x-1) 2004

    =(x-1) 2(-1+x) 2-x(x-1) 3+…-x(x-1) 2003+x(x-1) 2004

    =(x-1) 2(1-x)+…-x(x-1) 2003+x(x-1) 2004

    =(x-1) 2005