Đáp án:
a)
$\begin{array}{l}
Đkxđ:y \ne 0\\
\frac{{1 + 2013x}}{{60}} = \frac{{1 + 2015x}}{{5y}} = \frac{{1 + 2017x}}{{4y}} = \frac{{2 + 4030x}}{{10y}}\\
= \frac{{1 + 2013x + 1 + 2017x - \left( {2 + 4030x} \right)}}{{60 + 4y - 10y}} = 0\\
\Rightarrow \left\{ \begin{array}{l}
1 + 2013x = 0\\
1 + 2015x = 0\\
1 + 2017x = 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = - \frac{1}{{2013}}\\
x = - \frac{1}{{2015}}\\
x = - \frac{1}{{2017}}
\end{array} \right.
\end{array}$
Vậy không có giá trị của x và y thỏa mãn điều kiện
b)
$\begin{array}{l}
\frac{{x + y - 2016z}}{z} = \frac{{y + z - 2016x}}{x} = \frac{{z + x - 2016y}}{y}\\
= \frac{{x + y - 2016z + y + z - 2016x + z + x - 2016y}}{{z + x + y}}\\
= \frac{{ - 2014\left( {x + y + z} \right)}}{{x + y + z}} = - 2014\\
\Rightarrow \left\{ \begin{array}{l}
x + y - 2016z = - 2014z\\
y + z - 2016x = - 2014x\\
z + x - 2016y = - 2014y
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x + y = 2z\\
y + z = 2x\\
z + x = 2y
\end{array} \right. \Rightarrow x = y = z\\
C = \left( {1 + \frac{x}{y}} \right)\left( {1 + \frac{y}{z}} \right)\left( {1 + \frac{z}{x}} \right)\\
C = \left( {1 + 1} \right)\left( {1 + 1} \right)\left( {1 + 1} \right)\\
C = 8
\end{array}$
