Đáp án:
3) x=3
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \pm 2\\
A = \dfrac{x}{{x + 2}} + \dfrac{4}{{x - 2}} + \dfrac{{{x^2} - 3x + 18}}{{4 - {x^2}}}\\
= \dfrac{{x\left( {x - 2} \right) + 4\left( {x + 2} \right) - {x^2} + 3x - 18}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{{x^2} - 2x + 4x + 8 - {x^2} + 3x - 18}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{5x - 10}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{5}{{x + 2}}\\
2)A < 0\\
\to \dfrac{5}{{x + 2}} < 0\\
\to x + 2 < 0\\
\to x < - 2\\
3)A \in Z\\
\Leftrightarrow \dfrac{5}{{x + 2}} \in Z\\
\Leftrightarrow x + 2 \in U\left( 5 \right)\\
\to \left[ \begin{array}{l}
x + 2 = 5\\
x + 2 = - 5\\
x + 2 = 1\\
x + 2 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = - 7\left( l \right)\\
x = - 1\left( l \right)\\
x = - 3\left( l \right)
\end{array} \right.
\end{array}\)
