BPT `<=> x^3 - x^2 + 1 > m(x^2 - x)`

Vì `x in (1; 3) => x^2 - x > 0`

`=> m < x + 1/(x^2 - x), AA x in (1; 3)`

Xét `g(x) = x + 1/(x^2 - x)` trên `(1; 3)`

`g'(x) = 1 - (2x - 1)/(x^2 - x)^2 = (x^4 - 2x^3 + x^2 - 2x + 1)/(x^2 - x)^2`

`g'(x) = 0 <=> x^2 - 2x + 1 - 2/x + 1/x^2 = 0`

`<=> (x + 1/x)^2 - 2(x + 1/x) - 1 = 0`

Do `x in (1; 3) => x + 1/x > 2`

`=> x + 1/x = 1 + sqrt(2)`

`=> x^2 - (1 + sqrt(2))x + 1 = 0`

`=>EE` nghiệm `x_0=(1 + sqrt(2) + sqrt(2sqrt(2) - 1))/2 in (1;3)`

BBT: Xem ảnh

`=> "YCBT"<=> m < min_{(1; 3)} f(x) = f(x_0)`

`=> m < (-1 + sqrt(13 + 16sqrt(2)))/2`