\(\frac{{{x^2} - 9x + 14}}{{{x^2} - 5x + 4}} > 0\)
Giải chi tiết:
ĐKXĐ: \({x^2} - 5x + 4 \ne 0 \Leftrightarrow \left\{ \begin{array}{l}x \ne 4\\x \ne 1\end{array} \right.\)
Ta có: \({x^2} - 9x + 14 = 0 \Leftrightarrow \left( {x - 2} \right)\left( {x - 7} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}x = 2\\x = 7\end{array} \right..\)
\({x^2} - 5x + 4 = 0 \Leftrightarrow \left( {x - 1} \right)\left( {x - 4} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}x = 1\\x = 4\end{array} \right..\)
Đặt \(f\left( x \right) = \frac{{{x^2} - 9x + 14}}{{{x^2} - 5x + 4}}\) . Ta có bảng:

Vậy \(f\left( x \right) > 0 \Leftrightarrow x \in \left( { - \infty ;1} \right) \cup \left( {2;4} \right) \cup \left( {7; + \infty } \right).\)
Chọn D.