Đáp án`+`Giải thích các bước giải:

 `(x + 2)(x + 3) \ge 0`

`<=>` $\left[\begin{matrix} \begin{cases} x + 2 \ge 0\\x + 3\ge 0\end{cases}\\\begin{cases} x + 2 \le 0\\x + 3 \le 0\end{cases}\end{matrix}\right.$ 

`<=>` $\left[\begin{matrix} \begin{cases} x \ge -2\\x \ge -3\end{cases}\\\begin{cases} x \le -2\\x \le -3\end{cases}\end{matrix}\right.$ 

`<=>` $\left[\begin{matrix} x \ge -2\\x \le -3\end{matrix}\right.$ 

Vậy ..............

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`(2 -x)(x + 3) \ge 0`

`<=>` $\left[\begin{matrix} \begin{cases} 2 -x \ge 0\\x + 3\ge 0\end{cases}\\\begin{cases} 2 -x \le 0\\x +3 \le 0\end{cases}\end{matrix}\right.$ 

`<=>` $\left[\begin{matrix} \begin{cases} x \le 2\\x \ge -3\end{cases}\\\begin{cases} x \ge 2\\x \le -3\end{cases}\end{matrix}\right.$ 

`<=>` $\left[\begin{matrix} -3 \le x \le 2\\2 \le x \le - 3 (\text{vô lí})\end{matrix}\right.$ 
`<=> -3 \le x \le 2`

Vậy ........

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`x^2 + 5x +6 \ge 0`

`<=> x^2 + 2x + 3x +6 \ge 0`

`<=> x(x + 2) + 3(x + 2) \ge 0`

`<=> (x + 2)(x +3) \ge 0`

`<=>` $\left[\begin{matrix} \begin{cases} x + 2 \ge 0\\x + 3\ge 0\end{cases}\\\begin{cases} x + 2 \le 0\\x + 3 \le 0\end{cases}\end{matrix}\right.$ 

`<=>` $\left[\begin{matrix} \begin{cases} x \ge -2\\x \ge -3\end{cases}\\\begin{cases} x \le -2\\x \le -3\end{cases}\end{matrix}\right.$

 `<=>` $\left[\begin{matrix} x \ge -2\\x \le -3\end{matrix}\right.$ 

Vậy ..........

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`-x^2+ 5x - 6 > 0`

`<=> x^2 - 5x + 6 < 0`

`<=> x^2 - 2x - 3x + 6 < 0`

`<=> x(x - 2) - 3(x - 2) < 0`

`<=> (x - 2)(x - 3)  < 0`

`<=>`  $\left[\begin{matrix} \begin{cases} x - 2 < 0\\x -3 > 0\end{cases}\\\begin{cases} x -2 > 0\\x - 3< 0\end{cases}\end{matrix}\right.$ 

`<=>` $\left[\begin{matrix} \begin{cases} x < 2\\x >3\end{cases}\\\begin{cases} x > 2\\x <3 \end{cases}\end{matrix}\right.$ 

`<=>`  $\left[\begin{matrix} 3< x< 2 (\text{vô lí})\\2 < x < 3\end{matrix}\right.$ 

`<=> 2< x < 3`

Vậy ....

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`-x^2 + 5x - 6 \ge 0`

`<=> x^2 - 5x + 6 \le 0`

`<=> x^2 -2x - 3x + 6 \le 0`

`<=> x(x - 2) - 3(x - 2) \le 0`

`<=> (x - 2)(x - 3) \le 0`

`<=>`  $\left[\begin{matrix} \begin{cases} x - 2 \le 0\\x -3 \ge 0\end{cases}\\\begin{cases} x -2 \ge 0\\x - 3 \le0\end{cases}\end{matrix}\right.$ 

`<=>` $\left[\begin{matrix} \begin{cases} x\le2\\x \ge3\end{cases}\\\begin{cases} x\ge2\\x \le3 \end{cases}\end{matrix}\right.$

`<=>`  $\left[\begin{matrix} 3\le x\le2 (\text{vô lí})\\2 \le x \le 3\end{matrix}\right.$ 

`<=>2 \le x \le 3`

Vậy ..............

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`3 - x^2 + 5x- 6 < 0`

`<=> -x^2 + 5x - 3 < 0`

`<=> x^2 - 5x + 3 >0`

`<=> x^2 - 2.x. 5/2 + (5/2)^2 - (13)/4 < 0`

`<=> (x - 5/2)^2 - ((\sqrt{13})/2) < 0`

`<=> (x - (5 +\sqrt{13})/2)(x - (5 -\sqrt{13})/2) < 0`

`<=>`  $\left[\begin{matrix} \begin{cases} x - \dfrac{5 + \sqrt{13}}{2} < 0\\x - \dfrac{5 - \sqrt{13}}{2} >0\end{cases}\\\begin{cases}x - \dfrac{5 + \sqrt{13}}{2} > 0 \\x - \dfrac{5 - \sqrt{13}}{2} < 0 \end{cases}\end{matrix}\right.$ 

`<=>`   $\left[\begin{matrix} \begin{cases} x< \dfrac{5 + \sqrt{13}}{2} \\x > \dfrac{5 - \sqrt{13}}{2} \end{cases}\\\begin{cases}x > \dfrac{5 + \sqrt{13}}{2} \\x < \dfrac{5 - \sqrt{13}}{2} 0 \end{cases}\end{matrix}\right.$ 

`<=>` $\left[\begin{matrix} \dfrac{5 -\sqrt{13}}{2} < x <\dfrac{5 + \sqrt{13}}{2}\\ \dfrac{5 - \sqrt{13}}{2} > x > \dfrac{5 + \sqrt{13}}{2} (\text{vô lí})\end{matrix}\right.$

`<=> (5 - \sqrt{13})/2 < x < (5 +\sqrt{13})/2`

Vậy ....

`***`sharksosad