\(a,B=\left(\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}\right)\left(1-\dfrac{2}{x+3}\right)\)
\(\Rightarrow B=\left(\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{6x}{x^2-9}+\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right).\dfrac{x+3-2}{x+3}\)
\(\Rightarrow B=\left(\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}+\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x^2-3x}{\left(x-3\right)\left(x+3\right)}\right).\dfrac{x+1}{x+3}\)
\(\Rightarrow B=\dfrac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+1}{x+3}\)
\(\Rightarrow B=\dfrac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+1}{x+3}\)
\(\Rightarrow B=\dfrac{\left(x+3\right)^2\left(x+1\right)}{\left(x-3\right)\left(x+3\right)^2}\)
\(\Rightarrow B=\dfrac{x+1}{x-3}\)
\(b,B=\dfrac{1}{x}\Leftrightarrow\dfrac{x+1}{x-3}=\dfrac{1}{x}\\ \Leftrightarrow x\left(x+1\right)=x-3\\ \Leftrightarrow x^2+x=x-3\\ \Leftrightarrow x^2+3=0\left(vô.lí\right)\)
\(c,\) Để\(B>0\Leftrightarrow\dfrac{x+1}{x-3}>0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\\x-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1< 0\\x-3< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-1\\x>3\end{matrix}\right.\\\left\{{}\begin{matrix}x< -1\\x< 3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -1\end{matrix}\right.\)
