Help meeeeeeeeeeeeeeeeeeee

Đáp án:

`a,`

`(3x^2 + 5x + 4).(3x^2 - 9x + 4) + 24x^2`

`= 9x^4 - 27x^3 + 12x^2 + 15x^3 - 45x^2 + 20x + 12x^2 - 36x + 16 + 24x^2`

`= 9x^4 - (27x^3 - 15x^3) + (12x^2 - 45x^2 + 12x^2 + 24x^2) + (20x - 36x) + 16`

`= 9x^4 - 12x^3 + 3x^2 - 16x + 16`

`= 9x^4- 9x^3 - 3x^3 + 3x^2 - 16x + 16`

`= 9x^3.(x- 1) - 3x^2.(x - 1) - 16.(x - 1)`

`= (x - 1).(9x^3 - 3x^2 - 16)`

`= (x - 1).(9x^3 - 12x^2 + 9x^2 - 12x + 12x - 16)`

`= (x - 1).[3x^2.(3x - 4) + 3x.(3x - 4) + 4.(3x - 4)]`

`= (x - 1).(3x - 4).(3x^2 + 3x + 4)`

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`b,`

`A = (3x^2 + 5x + 4).(3x^2 - 9x + 4) + 24x^2`

`A =  (x - 1).(3x - 4).(3x^2 + 3x + 4)`

Để `A >= 0 `

`<=> (x - 1).(3x - 4).(3x^2 + 3x + 4) >= 0`

mà: `3x^2 + 3x + 4 = 3.(x^2 + x + 4/3) = 3.(x^2 + x + 1/4 + 13/12) = 3.(x + 1/2)^2 + 13/4 > 0`

`=> A >= 0 `

`<=> (x - 1).(3x - 4) >= 0`

$⇔ \left[\begin{matrix} \begin{cases} x - 1 \ge 0\\3x - 4 \ge 0 \end{cases}\\ \begin{cases} x - 1 \le 0\\3x- 4 \le 0\end{cases}\end{matrix}\right.⇔ \left[\begin{matrix} \begin{cases} x \ge 1\\x \ge \dfrac43 \end{cases}\\\begin{cases} x \le 1\\ x \le \dfrac43\end{cases}\end{matrix}\right. ⇔\left[\begin{matrix} x \ge \dfrac43 \\ x \le 1\end{matrix}\right.$

Vậy với `x <= 1` hoặc `x >= 4/3` thì `A >= 0`

`#dariana`