2x∧3-2x∧2-5x+2≤0怎样解?

1个回答

  • 2x³-2x²-5x+2≤0,

    2x³-2x²-5x+2

    =x³-2x²+x³-4x-(x-2)

    =x²(x-2)+x(x²-4)-(x-2)

    =x²(x-2)+x(x-2)(x+2)-(x-2)

    =(x-2)(x²+x(x+2)-1)

    =(x-2)(2x²+2x-1)

    =(x-2)[x+(1+√3)/2][x-(1-√3)/2]

    2x³-2x²-5x+2=(x-2)[x+(1+√3)/2][x-(1-√3)/2]≤0,

    得到以下四个不等式组

    一组:(x-2)≤0

    [x+(1+√3)/2]≤0

    [x-(1-√3)/2]≤0

    二组: (x-2)≤0

    [x+(1+√3)/2]≥0

    [x-(1-√3)/2]≥0

    三组:(x-2)≥0

    [x+(1+√3)/2)≤0

    [x-(1-√3)/2]≥0

    四组:(x-2)≥0

    [x+(1+√3)/2]≥0

    [x-(1-√3)/2]≤0

    由一组得:x≤2,x≤-(1+√3)/2,x≤(1-√3)/2]

    ∴x≤-(1+√3)/2,

    由二组得; (1-√3)/2 ≤x≤2;

    由三组得:

    x≥2,或x≤-(1+√3)/2

    由四组得:x≥2,或x≤-(1-√3)/2);

    所以综合上述得2x³-2x²-5x+2≤0,的解集为:

    (1-√3)/2 ≤x≤2;或(x≥2,或x≤-(1+√3)/2)