a. Để P được xđ thì MT phải khác 0.
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-9\ne0\\x^2+3x\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)\left(x+3\right)\ne0\\x\left(x+3\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\ne0\end{matrix}\right.\)
b. \(P=\left(\dfrac{x+9}{x^2-9}-\dfrac{3}{x^2+3x}\right).\dfrac{x-3}{x+3}\)
\(P=\left(\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\right).\dfrac{x-3}{x+3}\)
\(P=\left(\dfrac{x\left(x+9\right)}{x\left(x-3\right)\left(x+3\right)}-\dfrac{3\left(x-3\right)}{x\left(x+3\right)\left(x-3\right)}\right).\dfrac{x-3}{x+3}\)
\(P=\dfrac{x^2+9x-3x+9}{x\left(x-3\right)\left(x+3\right)}.\dfrac{x-3}{x+3}\)
\(P=\dfrac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}.\dfrac{x-3}{x+3}\)
\(P=\dfrac{1}{x}\)
