Đáp án:
a) \(\dfrac{3}{{x + 5}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)P = \left( {\dfrac{2}{{x + 4}} + \dfrac{{x + 20}}{{{x^2} - 16}}} \right).\dfrac{{x - 4}}{{x + 5}}\\
= \left[ {\dfrac{{2\left( {x - 4} \right) + x + 20}}{{\left( {x - 4} \right)\left( {x + 4} \right)}}} \right].\dfrac{{x - 4}}{{x + 5}}\\
= \dfrac{{3x + 12}}{{\left( {x - 4} \right)\left( {x + 4} \right)}}.\dfrac{{x - 4}}{{x + 5}}\\
= \dfrac{3}{{x - 4}}.\dfrac{{x - 4}}{{x + 5}}\\
= \dfrac{3}{{x + 5}}\\
b){x^2} + 4x = 0\\
\to x\left( {x + 4} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = - 4\left( l \right)
\end{array} \right.\\
Thay:x = 0\\
\to P = \dfrac{3}{5}\\
c)P \in Z\\
\Leftrightarrow \dfrac{3}{{x + 5}} \in Z\\
\Leftrightarrow x + 5 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
x + 5 = 3\\
x + 5 = - 3\\
x + 5 = 1\\
x + 5 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = - 2\\
x = - 8\\
x = - 4\left( l \right)\\
x = - 6
\end{array} \right.
\end{array}\)
