Đáp án:
$\begin{array}{l}
B = \left( {\frac{1}{3} + \frac{3}{{{x^2} - 3x}}} \right):\left( {\frac{{{x^2}}}{{27 - 3{x^2}}} + \frac{1}{{3 + x}}} \right)\\
a)Đkxd:\left\{ \begin{array}{l}
{x^2} - 3x \ne 0\\
27 - 3{x^2} \ne 0\\
3 + x \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x\left( {x - 3} \right) \ne 0\\
3\left( {9 - {x^2}} \right) \ne 0\\
x \ne - 3
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne 0;x \ne 3\\
x \ne 3;x \ne - 3\\
x \ne - 3
\end{array} \right. \Rightarrow x \ne - 3;x \ne 0;x \ne 3\\
b)B = \left( {\frac{1}{3} + \frac{3}{{{x^2} - 3x}}} \right):\left( {\frac{{{x^2}}}{{27 - 3{x^2}}} + \frac{1}{{3 + x}}} \right)\\
= \left( {\frac{1}{3} + \frac{3}{{x\left( {x - 3} \right)}}} \right):\left( {\frac{{{x^2}}}{{3\left( {3 - x} \right)\left( {3 + x} \right)}} + \frac{1}{{3 + x}}} \right)\\
= \frac{{{x^2} - 3x + 9}}{{3x\left( {x - 3} \right)}}:\left( {\frac{{{x^2} + 3\left( {3 - x} \right)}}{{3\left( {3 - x} \right)\left( {3 + x} \right)}}} \right)\\
= \frac{{{x^2} - 3x + 9}}{{3x\left( {x - 3} \right)}}.\frac{{3\left( {3 - x} \right)\left( {3 + x} \right)}}{{{x^2} - 3x + 9}}\\
= \frac{{ - x - 3}}{x} = - 1 - \frac{3}{x}\\
c)B \in Z\\
\Rightarrow - 1 - \frac{3}{x} \in Z\\
\Rightarrow \frac{3}{x} \in Z\\
\Rightarrow x \in Ư\left( 3 \right) = {\rm{\{ }}1;3\}
\end{array}$
Vậy x=1 hoặc x=3 thì B nguyên
