整式乘除与因式分解的初一题1.化简(1)(2a+1)(-a-2)+a(2-a)(2)[(x+y)²-(x+y)

4个回答

  • 1. 化简

    (1)

    (2a+1)(-a-2)+a(2-a)

    = (-2a² -5a -2) +(2a -a²)

    =-3a²-3a-2

    (2)

    [(x+y)²-(x+y)(x-y)]÷(-2-y)

    = [(x² +2xy +y²) -(x² -y²)] ÷ [-(y+2)]

    = -[2y² +2xy] ÷ (y+2)

    = -(2y² +2xy)/(y+2)

    =2xy/-y-2

    2. 分解因式

    -2x³+8xy²

    = -2x(x² -4y²)

    = -2x(x+2y)(x-2y)

    3.已知x²-2x=2,求代数式(x-1)²+(x+3)(x-3)+(x-3)(x-1)的值

    (x-1)²+(x+3)(x-3)+(x-3)(x-1)

    = (x² -2x +1) +(x² -9) +(x² -4x +3)

    = 3x² -6x -5

    = 3(x² -2) -5

    当x²-2x=2时

    原式

    = 3*2 -5

    = 1.