Tìm `x` để `A` > `0` `A` `=` $\frac{-3(x-3)}{x+3}$

$\text{Có: A > 0 Đkxđ: x $\neq$ -3}$

$\text{⇔ $\dfrac{-3(x - 3)}{x + 3}$ > 0}$

$\text{⇔ $\dfrac{x - 3}{x + 3}$ < 0}$

$\text{⇔ $\left[\begin{matrix} \begin{cases} x - 3 > 0\\x + 3 < 0 \end{cases}\\ \begin{cases} x - 3 < 0\\ x + 3 > 0 \end{cases}\end{matrix}\right.$}$

$\text{⇔ $\left[\begin{matrix} \begin{cases} x > 3\\x < -3 (Vô lí) \end{cases}\\ \begin{cases} x < 3\\x > -3\end{cases}\end{matrix}\right.$}$

$\text{⇔ -3 < x < 3}$

$\text{Vậy - 3 < x < 3 thì A > 0}$

$\textit{Ha1zzz}$