计算:(B-1)(1+B)(B²-1),(a-2b+3c)(a+2b-3c)

1个回答

  • 计算:

    1.(B-1)(1+B)(B²-1)

    =(B-1)(B+1)(B²-1)

    =(B²-1)(B²-1)

    =(B²-1)²

    2.(a-2b+3c)(a+2b-3c)

    =[a-(2b-3c)][a+(2b-3c)]

    =a² -(2b-3c)²

    分解因式:

    1.a²x²+16ax +64

    =(ax)²+2*8*ax +8²

    =(ax+8)²

    2.25(x+y)²-16(x-y)²

    =[5*(x+y)]²-[4(x-y)]²

    =[5(x+y)+4(x-y)]*[(5(x+y)-4(x-y)]

    =(9x-y)*(x+9y)

    =9x²+81xy-xy-9²

    =9x²+80xy-9y²

    3.x²-6x+9-Y²

    =x²-2*3x+3*3-Y²

    =(x-3)²-Y²

    =(x-3-y)*(x-3+y)

    4.(a²+b²-1)²-4a²b²

    =(a²+b²-1)²-(2ab)²

    =[(a²+b²-1)+2ab]*[(a²+b²-1)-2ab]

    =[(a+b)²-1]*[(a-b)²-1]

    求值:

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

    =[(x²-(2y- 3/2)²]*(-3x)+[4y²-6y+ (√2)/4]*(-3x)

    将X+-1,Y=2007/2006代入即可