Đáp án:

\(\,m \in \left\{ {\frac{3}{2};\,\,3} \right\}\)

Giải thích các bước giải: \(\begin{array}{l} \left( {m - 2} \right){x^2} - \left( {m - 4} \right)x - 2 = 0\,\,\,\left( * \right)\\ Phuong\,\,trinh\,\,\left( * \right)\,\,\,co\,\,hai\,\,nghiem\,\,\,phan\,\,biet\,\,\,{x_1},\,\,{x_2}\\ \Leftrightarrow \left\{ \begin{array}{l} m - 2 \ne 0\\ \Delta > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m \ne 2\\ {\left( {m - 4} \right)^2} + 8\left( {m - 2} \right) > 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} m \ne 2\\ {m^2} - 8m + 16 + 8m - 16 > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m \ne 2\\ {m^2} > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m \ne 2\\ m \ne 0 \end{array} \right..\\ Ap\,\,dung\,\,he\,\,thuc\,\,Vi - et\,\,ta\,\,co:\,\,\,\left\{ \begin{array}{l} {x_1} + {x_2} = \frac{{m - 4}}{{m - 2}}\\ {x_1}{x_2} = \frac{{ - 2}}{{m - 2}} \end{array} \right.\\ Lai\,\,co:\,\,{x_1} - {x_2} = 3 \Leftrightarrow {x_2} = {x_1} - 3\\ \Rightarrow 2{x_1} = \frac{{m - 4}}{{m - 2}} + 3 = \frac{{m - 4 + 3m - 6}}{{m - 2}}\\ \Leftrightarrow 2{x_1} = \frac{{4m - 10}}{{m - 2}} \Leftrightarrow {x_1} = \frac{{2m - 5}}{{m - 2}}\\ \Rightarrow {x_2} = \frac{{2m - 5}}{{m - 2}} - 3 = \frac{{2m - 5 - 3m + 6}}{{m - 2}} = \frac{{ - m + 1}}{{m - 2}}\\ \Rightarrow {x_1}{x_2} = \frac{{ - 2}}{{m - 2}}\\ \Leftrightarrow \frac{{2m - 5}}{{m - 2}}.\frac{{ - m + 1}}{{m - 2}} = \frac{{ - 2}}{{m - 2}}\\ \Leftrightarrow \left( {2m - 5} \right)\left( {m - 1} \right) = 2\left( {m - 2} \right)\\ \Leftrightarrow 2{m^2} - 7m + 5 - 2m + 4 = 0\\ \Leftrightarrow 2{m^2} - 9m + 9 = 0\\ \Leftrightarrow \left( {2m - 3} \right)\left( {m - 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} 2m - 3 = 0\\ m - 3 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} m = \frac{3}{2}\,\,\left( {tm} \right)\\ m = 3\,\,\,\left( {tm} \right) \end{array} \right.\\ Vay\,\,m \in \left\{ {\frac{3}{2};\,\,3} \right\}. \end{array}\)