2x∧3-2x∧2-5x+2≤0怎样解?

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2x∧3-2x∧2-5x+2≤0怎样解?
2x∧3-2x∧2-5x+2≤0怎样解?

2x∧3-2x∧2-5x+2≤0怎样解?
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)