Giải thích các bước giải:
$\begin{array}{l} a)\dfrac{6}{{2x - 3}} \in Z\\ \Rightarrow 6 \vdots \left( {2x - 3} \right)\\ \Rightarrow \left( {2x - 3} \right) \in \left\{ { - 6; - 3; - 2; - 1;1;2;3;6} \right\}\\ \Rightarrow 2x \in \left\{ { - 3;0;1;2;4;5;6;9} \right\}\\ \Rightarrow x \in \left\{ { - \dfrac{3}{2};0;\dfrac{1}{2};1;2;\dfrac{5}{2};3;\dfrac{9}{2}} \right\}\\ Do:x \in Z\\ \Rightarrow x \in \left\{ {0;1;2;3} \right\}\\ b)\dfrac{{2x - 3}}{{x + 1}} = \dfrac{{2x + 2 - 5}}{{x + 1}} = \dfrac{{2\left( {x + 1} \right) - 5}}{{x + 1}}\\ = 2 - \dfrac{5}{{x + 1}}\\ Do:2 \in Z\\ \Rightarrow \dfrac{5}{{x + 1}} \in Z\\ \Rightarrow \left( {x + 1} \right) \in \left\{ { - 5; - 1;1;5} \right\}\\ \Rightarrow x \in \left\{ { - 6; - 2;0;4} \right\}\\ \text{Vậy}\,x \in \left\{ { - 6; - 2;0;4} \right\}\\ c)\dfrac{{2x - 3}}{{3x - 2}} \in Z\\ \Rightarrow \left( {2x - 3} \right) \vdots \left( {3x - 2} \right)\\ \Rightarrow 3.\left( {2x - 3} \right) \vdots \left( {3x - 2} \right)\\ Do:3\left( {2x - 3} \right) = 6x - 9\\ = 6x - 4 - 5\\ = 2.\left( {3x - 2} \right) - 5\\ Do:2\left( {3x - 2} \right) \vdots \left( {3x - 2} \right)\\ \Rightarrow 5 \vdots \left( {3x - 2} \right)\\ \Rightarrow \left( {3x - 2} \right) \in \left\{ { - 5; - 1;1;5} \right\}\\ \Rightarrow 3x \in \left\{ { - 3;1;3;7} \right\}\\ \Rightarrow x \in \left\{ { - 1;\dfrac{1}{3};1;\dfrac{7}{3}} \right\}\\ Do:x \in Z\\ \Rightarrow x \in \left\{ { - 1;1} \right\}\\ \text{Vậy}\,x \in \left\{ { - 1;1} \right\} \end{array}$
