Question

(x^3-2x+1)(x^3+2x-3)≤0 怎么解

Answer 1

解：
(x³-2x+1)(x³+2x-3)≤0
(x³-x²+x²-x-x+1)(x³-x²+x²-x+3x-3)≤0
[x²(x-1)+x(x-1)-(x-1)][x²(x-1)+x(x-1)+3(x-1)]≤0
(x-1)(x²+x-1)(x-1)(x²+x+3)≤0
(x-1)²(x²+x+3)(x²+x+¼ -5/4)≤0
(x-1)²(x²+x+3)[(x+½)² -5/4]≤0
(x-1)²(x²+x+3)[x+(1+√5)/2][x+(1-√5)/2]≤0
x²+x+3=x²+x+¼+ 11/4=(x+½)²+ 11/4恒>0
(x-1)²恒≥0
因此只有[x+(1+√5)/2][x+(1-√5)/2]≤0
-(1+√5)/2≤x≤(√5-1)/2
不等式的解集为[-(1+√5)/2，(√5-1)/2]

Answer 2

let
f(x) = x^3+2x-3
f(1) = 0
x^3+2x-3 = (x-1)(x^2+ax+3)
coef. of x
3-a=2
a=1
x^3+2x-3 = (x-1)(x^2+x+3)
= (x-1)[(x^2+1/2)^2+11/4]

(x^3-2x+1)(x^3+2x-3)≤0
(x-1)^2. (x^3+2x-3)≤0
(x^3+2x-3)≤0
(x-1)[(x^2+1/2)^2+11/4]≤0
x-1≤0
x≤1