Giải thích các bước giải:
a.Để $A$ xác định
$\to\begin{cases}x+3\ne 0\\ x-3\ne 0\\ 9-x^2\ne 0\end{cases}$
$\to\begin{cases}x\ne -3\\ x\ne 3\\ x^2\ne 9\to x\ne\pm3\end{cases}$
$\to x\ne \pm3$
b.Ta có:
$A=\dfrac{3}{x+3}+\dfrac{1}{x-3}-\dfrac{18}{9-x^2}$
$\to A=\dfrac{3}{x+3}+\dfrac{1}{x-3}+\dfrac{18}{x^2-9}$
$\to A=\dfrac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{x+3}{\left(x+3\right)\left(x-3\right)}+\dfrac{18}{\left(x+3\right)\left(x-3\right)}$
$\to A=\dfrac{3\left(x-3\right)+x+3+18}{\left(x+3\right)\left(x-3\right)}$
$\to A=\dfrac{4x+12}{\left(x+3\right)\left(x-3\right)}$
$\to A=\dfrac{4\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}$
$\to A=\dfrac{4}{x-3}$
c.Để $A=4$
$\to \dfrac4{x-3}=4$
$\to x-3=1$
$\to x=4$
