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