Đáp án:

\[m > 2\sqrt 2 - 2\]

Giải thích các bước giải: \[\begin{array}{l} Cau\,\,2:\\ y = \frac{{2{x^2} + mx + 2 - m}}{{x + m - 1}}\,\,\,DB\,\,\,tren\,\,\,\left( {1; + \infty } \right)\\ DK:\,\,\,x \ne 1 - m\\ \Rightarrow y' = \frac{{\left( {4x + m} \right)\left( {x + m - 1} \right) - \left( {2{x^2} + mx + 2 - m} \right)}}{{{{\left( {x + m - 1} \right)}^2}}}\\ = \frac{{4{x^2} + 4\left( {m - 1} \right)x + mx + {m^2} - m - 2{x^2} - mx - 2 + m}}{{{{\left( {x + m - 1} \right)}^2}}}\\ = \frac{{2{x^2} + 4\left( {m - 1} \right)x + {m^2} - 2}}{{{{\left( {x + m - 1} \right)}^2}}}.\\ \Rightarrow y' = 0\\ \Leftrightarrow 2{x^2} + 4\left( {m - 1} \right)x + {m^2} - 2 = 0\\ \Delta ' = 4{\left( {m - 1} \right)^2} - 2{m^2} + 4 = 4{m^2} - 8m + 4 - 2{m^2} + 4\\ = 2{m^2} - 8m + 8 = 2\left( {{m^2} - 4m + 4} \right) = 2{\left( {m - 2} \right)^2}.\\ \Rightarrow \Delta ' > 0 \Leftrightarrow m \ne 2\\ Bang\,\,xet\,\,dau:\\ \,\,\,\,\, + \,\,\,\,\,\,\,\,{x_1}\,\,\,\,\, - \,\,\,\,\,\,\,\,\,{x_2}\,\,\,\,\,\, + \\ \Rightarrow hs\,\,\,DB\,\,\,tren\,\,\,\left( {1;\,\, + \infty } \right)\\ \Leftrightarrow \left\{ \begin{array}{l} {x_1} < {x_2} < 1\\ 1 - m \notin \left( {1; + \infty } \right) \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {x_1} + {x_2} < 2\\ \left( {{x_1} - 1} \right)\left( {{x_2} - 1} \right) > 0\\ 1 - m \le 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} - 2\left( {m - 1} \right) < 2\\ {x_1}{x_2} - \left( {{x_1} + {x_2}} \right) + 1 > 0\\ m \ge 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m - 1 > - 1\\ \frac{{{m^2} - 2}}{2} + 2\left( {m - 1} \right) + 1 > 0\\ m \ge 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} m > 0\\ m \ge 0\\ {m^2} - 2 + 4m - 4 + 2 > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m > 0\\ {m^2} + 4m - 4 > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m > 0\\ \left[ \begin{array}{l} m > - 2 + 2\sqrt 2 \\ m < - 2 - 2\sqrt 2 \end{array} \right. \end{array} \right. \Leftrightarrow m > 2\sqrt 2 - 2. \end{array}\]