a: Sửa đề: \(B=\dfrac{x}{x+3}+\dfrac{2x}{x-3}-\dfrac{9-3x^2}{x^2-9}\)
ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
\(B=\dfrac{x}{x+3}+\dfrac{2x}{x-3}-\dfrac{9-3x^2}{x^2-9}\)
\(=\dfrac{x}{x+3}+\dfrac{2x}{x-3}+\dfrac{3x^2-9}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x\left(x-3\right)+2x\left(x+3\right)+3x^2-9}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{x^2-3x+2x^2+6x+3x^2-9}{\left(x+3\right)\left(x-3\right)}=\dfrac{6x^2+3x-9}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{6x^2+9x-6x-9}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{\left(6x+9\right)\left(x-1\right)}{\left(x+3\right)\left(x-3\right)}\)
b:
Để B<0 thì \(\dfrac{\left(6x+9\right)\left(x-1\right)}{\left(x+3\right)\left(x-3\right)}< 0\)
TH1: \(\left\{{}\begin{matrix}\left(6x+9\right)\left(x-1\right)>0\\\left(x+3\right)\left(x-3\right)< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x< -\dfrac{3}{2}\end{matrix}\right.\\-3< x< 3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3< x< -\dfrac{3}{2}\\1< x< 3\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}\left(6x+9\right)\left(x-1\right)< 0\\\left(x+3\right)\left(x-3\right)>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{3}{2}< x< 1\\\left[{}\begin{matrix}x>3\\x< -3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Để B>0 thì \(\dfrac{\left(6x+9\right)\left(x-1\right)}{\left(x+3\right)\left(x-3\right)}>0\)
TH1: \(\left\{{}\begin{matrix}\left(6x+9\right)\left(x-1\right)>0\\\left(x+3\right)\left(x-3\right)>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x< -\dfrac{3}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x>3\\x< -3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -3\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}\left(6x+9\right)\left(x-1\right)< 0\\\left(x+3\right)\left(x-3\right)< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{3}{2}< x< 1\\-3< x< 3\end{matrix}\right.\Leftrightarrow-\dfrac{3}{2}< x< 1\)