Giải thích các bước giải:
$\begin{array}{l} Dkxd:\left\{ \begin{array}{l} x \ne 4\\ x \ne 3 \end{array} \right.\\ a)Q = \dfrac{{2x + 1}}{{{x^2} - 7x + 12}} - \dfrac{{x + 3}}{{x - 4}} - \dfrac{{2x + 1}}{{3 - x}}\\ = \dfrac{{2x + 1 - \left( {x + 3} \right)\left( {x - 3} \right) + \left( {2x + 1} \right)\left( {x - 4} \right)}}{{\left( {x - 3} \right)\left( {x - 4} \right)}}\\ = \dfrac{{2x + 1 - \left( {{x^2} - 9} \right) + 2{x^2} - 8x + x - 4}}{{\left( {x - 3} \right)\left( {x - 4} \right)}}\\ = \dfrac{{2x + 1 - {x^2} + 9 + 2{x^2} - 7x - 4}}{{\left( {x - 3} \right)\left( {x - 4} \right)}}\\ = \dfrac{{{x^2} - 5x + 6}}{{\left( {x - 3} \right)\left( {x - 4} \right)}}\\ = \dfrac{{\left( {x - 3} \right)\left( {x - 2} \right)}}{{\left( {x - 3} \right)\left( {x - 4} \right)}}\\ = \dfrac{{x - 2}}{{x - 4}}\\ b)\left| {2x - 1} \right| = 5\\ \Rightarrow \left[ \begin{array}{l} 2x - 1 = 5\\ 2x - 1 = - 5 \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} 2x = 6\\ 2x = - 4 \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} x = 3\left( {ktm} \right)\\ x = - 2\left( {tm} \right) \end{array} \right.\\ Khi:x = - 2\\ \Rightarrow Q = \dfrac{{x - 2}}{{x - 4}} = \dfrac{{ - 2 - 2}}{{ - 2 - 4}} = \dfrac{2}{3}\\ c)Q = 0\\ \Rightarrow \dfrac{{x - 2}}{{x - 4}} = 0\\ \Rightarrow x = 2\left( {tmdk} \right)\\ \text{Vậy}\,x = 2\\ d)Q = - 3\\ \Rightarrow \dfrac{{x - 2}}{{x - 4}} = - 3\\ \Rightarrow x - 2 = - 3\left( {x - 4} \right)\\ \Rightarrow x - 2 = - 3x + 12\\ \Rightarrow x + 3x = 12 + 2\\ \Rightarrow 4x = 14\\ = x = \dfrac{7}{2}\left( {tmdk} \right)\\ \text{Vậy}\,x = \dfrac{7}{2}\\ e)Q = \dfrac{{x - 2}}{{x - 4}} = \dfrac{{x - 4 + 2}}{{x - 4}}\\ = 1 + \dfrac{2}{{x - 4}}\\ Q \in Z\\ \Rightarrow \dfrac{2}{{x - 4}} \in Z\\ \Rightarrow 2 \vdots \left( {x - 4} \right)\\ \Rightarrow \left( {x - 4} \right) \in \left\{ { - 2; - 1;1;2} \right\}\\ \Rightarrow x \in \left\{ {2;3;5;6} \right\}\\ Do:x \ne 3;x \ne 4\\ \Rightarrow x \in \left\{ {2;5;6} \right\}\\ \text{Vậy}\,x \in \left\{ {2;5;6} \right\} \end{array}$
