Đáp án:
f. a>32
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:a \ne 17\\
X \in Q\\
\Leftrightarrow \dfrac{{15}}{{a - 17}} \in Q\\
\Leftrightarrow a - 17 \in U\left( {15} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
a - 17 = 15\\
a - 17 = - 15\\
a - 17 = 5\\
a - 17 = - 5\\
a - 17 = 3\\
a - 17 = - 3\\
a - 17 = 1\\
a - 17 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
a = 32\\
a = 2\\
a = 22\\
a = 12\\
a = 20\\
a = 14\\
a = 18\\
a = 16
\end{array} \right.\\
b.X \in {Q^ + }\\
\Leftrightarrow \left[ \begin{array}{l}
a - 17 = 15\\
a - 17 = 5\\
a - 17 = 3\\
a - 17 = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
a = 32\\
a = 22\\
a = 20\\
a = 18
\end{array} \right.\\
c.X \in {Q^ - }\\
\Leftrightarrow \left[ \begin{array}{l}
a - 17 = - 15\\
a - 17 = - 5\\
a - 17 = - 3\\
a - 17 = - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
a = 2\\
a = 12\\
a = 14\\
a = 16
\end{array} \right.\\
d.X = - 1\\
\to \dfrac{{15}}{{a - 17}} = - 1\\
\to 15 = - a + 17\\
\to - a = - 2\\
\to a = 2\\
e.X > 1\\
\to \dfrac{{15}}{{a - 17}} > 1\\
\to \dfrac{{15 - a + 17}}{{a - 17}} > 0\\
\to \dfrac{{32 - a}}{{a - 17}} > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
32 - a > 0\\
a - 17 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
32 - a < 0\\
a - 17 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
32 > a\\
a > 17
\end{array} \right.\\
\left\{ \begin{array}{l}
32 < a\\
a < 17
\end{array} \right.\left( l \right)
\end{array} \right.\\
f.0 < x < 1\\
\to \left\{ \begin{array}{l}
\dfrac{{15}}{{a - 17}} > 0\\
\dfrac{{15}}{{a - 17}} < 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
a - 17 > 0\\
\dfrac{{15 - a + 17}}{{a - 17}} < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
a - 17 > 0\\
32 - a < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
a > 17\\
a > 32
\end{array} \right.\\
\to a > 32
\end{array}\)
