Đáp án:
c) \(\left[ \begin{array}{l}
x = 1\\
x = - 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne 0\\
C = \dfrac{{x + 1}}{{2x}}\\
a)Thay:x = 2017\\
\to C = \dfrac{{2017 + 1}}{{2017.2}} = \dfrac{{2018}}{{2017.2}} = \dfrac{{1009}}{{2017}}\\
b)C > \dfrac{1}{2}\\
\to \dfrac{{x + 1}}{{2x}} > \dfrac{1}{2}\\
\to \dfrac{{2x + 2 - 2x}}{{4x}} > 0\\
\to \dfrac{2}{{4x}} > 0\\
\to x > 0\\
c)C = \dfrac{{x + 1}}{{2x}} \to 2C = \dfrac{{x + 1}}{x} = 1 + \dfrac{1}{x}\\
C \in Z\\
\Leftrightarrow \dfrac{1}{x} \in Z\\
\Leftrightarrow x \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 1
\end{array} \right.
\end{array}\)
