Điều kiện xác định $x\ne 0;5$
$\begin{array}{l}
\dfrac{{180}}{x} + \dfrac{{180}}{{x - 5}} + \dfrac{3}{2} = 10\\
 \Leftrightarrow 180\left( {\dfrac{1}{x} + \dfrac{1}{{x - 5}}} \right) = \dfrac{{17}}{2}\\
 \Leftrightarrow \dfrac{1}{x} + \dfrac{1}{{x - 5}} = \dfrac{{17}}{{360}}\\
 \Leftrightarrow \dfrac{{x - 5 + x}}{{x\left( {x - 5} \right)}} = \dfrac{{17}}{{360}}\\
 \Leftrightarrow \dfrac{{2x - 5}}{{x\left( {x - 5} \right)}} = \dfrac{{17}}{{360}}\\
 \Rightarrow 360\left( {2x - 5} \right) = 17x\left( {x - 5} \right)\\
 \Leftrightarrow 720x - 1800 = 17{x^2} - 85x\\
 \Leftrightarrow 17{x^2} - 805x + 1800 = 0\\
 \Leftrightarrow 17{x^2} - 40x - 765x + 1800 = 0\\
 \Leftrightarrow 17x\left( {x - \dfrac{{40}}{{17}}} \right) - 765\left( {x - \dfrac{{40}}{{17}}} \right) = 0\\
 \Leftrightarrow 17\left( {x - \dfrac{{40}}{{17}}} \right)\left( {x - 45} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{40}}{{17}}\\
x = 45
\end{array} \right.\left( {tm} \right)\\
 \Rightarrow S = \left\{ {\dfrac{{40}}{{17}};45} \right\}
\end{array}$