Giải thích các bước giải:
1.Ta có:
$A=\dfrac{3+x}{3-x}:\dfrac{9x^2}{x^2-6x+9}\cdot (\dfrac{3}{3-x}-\dfrac{9}{x^3+27}\cdot \dfrac{3x-9-x^2}{x-3})$
$\to A=\dfrac{3+x}{3-x}:\dfrac{9x^2}{(x-3)^2}\cdot (-\dfrac{3}{x-3}+\dfrac{9}{x^3+27}\cdot \dfrac{x^2-3x+9}{x-3})$
$\to A=\dfrac{3+x}{3-x}\cdot \dfrac{(x-3)^2}{9x^2}\cdot (-\dfrac{3}{x-3}+\dfrac{9}{x^3+27}\cdot \dfrac{x^2-3x+9}{x-3})$
$\to A=\dfrac{3+x}{3-x}\cdot \dfrac{(x-3)^2}{9x^2}\cdot (-\dfrac{3}{x-3}+\dfrac{9}{(x+3)(x^2-3x+9)}\cdot \dfrac{x^2-3x+9}{x-3})$
$\to A=-\dfrac{x+3}{x-3}\cdot \dfrac{\left(x-3\right)^2}{9x^2}\left(\dfrac{9}{\left(x+3\right)\left(x-3\right)}-\dfrac{3}{x-3}\right)$
$\to A=-\dfrac{x+3}{x-3}\cdot \dfrac{\left(x-3\right)^2}{9x^2}\left(-\dfrac{3x}{\left(x-3\right)\left(x+3\right)}\right)$
$\to A=\dfrac{3+x}{x-3}\cdot \dfrac{\left(x-3\right)^2}{9x^2}\cdot \dfrac{3x}{\left(x-3\right)\left(x+3\right)}$
$\to A=\dfrac{x}{3x^2}$
$\to A=\dfrac1{3x}$
2.Để $A$ xác định
$\to \begin{cases} 3-x\ne 0\\x\ne 0\\ x^2-6x+9\ne 0\\ x^3+27\ne 0\end{cases}$
$\to \begin{cases} x\ne 3\\x\ne 0\\ (x-3)\ne 0\\ x^3\ne -27\end{cases}$
$\to \begin{cases} x\ne 3\\x\ne 0\\ x\ne 3\\ x\ne -3\end{cases}$
$\to x\ne \pm3,0$
3.a.Khi $x=-\dfrac16$
$\to \dfrac1{3x}=-\dfrac16$
$\to 3x=-6$
$\to x=-2$
b.Để $A<0$
$\to \dfrac1{3x}<0$
$\to x<0$ vì $\dfrac13>0$
$\to x<0, x\ne -3$ vì $x\ne \pm3,0$
