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  • 分解因式: (1)(x2+4x+8)2+3x(x2+4x+8)+2x2; (2)(2x2-3x+1)2-22x2+33x-1; (3)x4+2001x2+2000x+2001; (4)(6x-1)(2 x-1)(3 x-1)( x-1)+x

    分解因式:
    (1)(x2+4x+8)2+3x(x2+4x+8)+2x2;
    (2)(2x2-3x+1)2-22x2+33x-1;
    (3)x4+2001x2+2000x+2001;
    (4)(6x-1)(2 x-1)(3 x-1)( x-1)+x2
    (5)a2+2b2+3c2+3ab+4ac+5bc;
    數(shù)學(xué)人氣:309 ℃時(shí)間:2020-03-31 06:18:30
    優(yōu)質(zhì)解答
    (1)(x2+4x+8)2+3x(x2+4x+8)+2x2
    =(x2+4x+8+x)(x2+4x+2x+8)
    =(x2+5x+8)(x+2)(x+4);
    (2)(2x2-3x+1)2-22x2+33x-1
    =(2x2-3x+1)2-11(2x2-3x)-1
    =(2x2-3x)2-9(2x2-3x)
    =(2x2-3x)(2x2-3x-9)
    =x(2x-3)(2x+3)(x-3);
    (3)x4+2001x2+2000x+2001
    =(x4+x3+x2)-(x3+x2+x)+(2001x2+2001x+2001)
    =x2(x2+x+1)-x(x2+x+1)+2001(x2+x+1)
    =(x2+x+1)(x2-x+2001);
    (4)(6x-1)(2 x-1)(3 x-1)( x-1)+x2
    =(6x-1)( x-1)(2 x-1)(3 x-1)+x2
    =(6x2-7x+1)(6x2-5x+1)+x2
    =(6x2+1-7x)(6x2+1-5x)+x2
    =(6x2+1)2-12x(6x2+1)+36x2
    =(6x2+1-6x)2;
    (5)a2+2b2+3c2+3ab+4ac+5bc
    =a2+3ab+4ac+2b2+5bc+3c2
    =a2+a(3b+4c)+(2b+3c)(b+c)
    =(a+2b+3c)(a+b+c).
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