Đáp án:
a) \(\dfrac{{ 2x + 4}}{{x - 3}}\)
b) \(\left[ \begin{array}{l}
x = 8\\
x = - 2\\
x = 4\\
x = 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \pm 3\\
A = \left( {\dfrac{x}{{x - 3}} - \dfrac{1}{{x + 3}} + \dfrac{{{x^2} - 1}}{{9 - {x^2}}}} \right):\dfrac{2}{{x + 3}}\\
= \left[ {\dfrac{{x\left( {x + 3} \right) - x + 3 - {x^2} + 1}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}} \right].\dfrac{{x + 3}}{2}\\
= \dfrac{{{x^2} + 3x - x - {x^2} + 4}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{x + 3}}{2}\\
= \dfrac{{2x + 4}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{x + 3}}{2}\\
= \dfrac{{x + 2}}{{x - 3}}\\
b)A = \dfrac{{x + 2}}{{x - 3}} = \dfrac{{x - 3 + 5}}{{x - 3}}\\
= 1 + \dfrac{5}{{x - 3}}\\
A \in Z \to \dfrac{5}{{x - 3}} \in Z\\
\to x - 3 \in U\left( 5 \right)\\
\to \left[ \begin{array}{l}
x - 3 = 5\\
x - 3 = - 5\\
x - 3 = 1\\
x - 3 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 8\\
x = - 2\\
x = 4\\
x = 2
\end{array} \right.
\end{array}\)
