Để \(A = \frac{{{x^2} - 6x + 8}}{{x - 1}} \le 0\) thì x thuộc:

A \(\left( { - \infty ;2} \right) \cup \left( {4; + \infty } \right)\) 

B \(\left[ {1;2} \right) \cup \left( {4; + \infty } \right)\) 

C \(\left( {2;3} \right)\)  

D \(\left( { - \infty ;1} \right) \cup \left[ {2;4} \right]\)