Giải thích các bước giải:
\(\begin{array}{l}
2)a)DK:x \ne \pm 3;x \ne 2\\
B = \dfrac{{{{\left( {x + 3} \right)}^2} - 2{x^2} + 6 + x\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{2\left( {x - 3} \right)\left( {x + 3} \right)}}{{6\left( {x - 2} \right)}}\\
= \dfrac{{{x^2} + 6x + 9 - 2{x^2} + 6 + {x^2} - 3x}}{{3\left( {x - 2} \right)}}\\
= \dfrac{{3x + 15}}{{3\left( {x - 2} \right)}} = \dfrac{{3\left( {x + 5} \right)}}{{3\left( {x - 2} \right)}} = \dfrac{{x + 5}}{{x - 2}}\\
b)\left| {x + 1} \right| = 2 \Leftrightarrow \left[ \begin{array}{l}
x = 1\left( {tm} \right)\\
x = - 3\left( {ktm} \right)
\end{array} \right.\\
B = \dfrac{{1 + 5}}{{1 - 2}} = - 6\\
c)B = \dfrac{{x + 5}}{{x - 2}} = \dfrac{{x - 2 + 7}}{{x - 2}} = 1 + \dfrac{7}{{x - 2}}\\
B \in Z \Rightarrow \dfrac{7}{{x - 2}} \in Z\\
\Rightarrow \left( {x - 2} \right) \in U\left( 7 \right) = \left\{ { - 1;1; - 7;7} \right\}\\
\Rightarrow x \in \left\{ {1;3; - 5;9} \right\}\\
KHDK \Rightarrow x \in \left\{ {1; - 5;9} \right\}
\end{array}\)
