\(f\left( x \right) = \left( {{m^2} - 3m + 2} \right){x^2} + 2(2 - m)x - 2\)

Cho f(x) = 0. Để f(x) có 2 nghiệm dương phân biệt.

\(\left\{ {\begin{array}{*{20}{c}}{a \ne 0}\\{\Delta ' > 0}\\{P > 0}\\{S > 0}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{{m^2} - 3m + 2 \ne 0}\\{{{\left( {2 - m} \right)}^2} - \left( {{m^2} - 3m + 2} \right)\left( { - 2} \right) > 0}\\{\frac{2}{{{m^2} - 3m + 2}} > 0}\\{\frac{{ - 4 + 2m}}{{{m^2} - 3m + 2}} > 0}\end{array}} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{m \ne 2;m \ne 1}\\{4 - 4m + {m^2} + 2{m^2} - 6m + 4 > 0}\\{{m^2} - 3m + 2 > 0}\\{1 < m < 2;m > 2}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{m \ne 2;m \ne 1}\\{3{m^2} - 10m + 8 > 0}\\{m < 1;m > 2}\\{1 < m < 2;m > 2}\end{array}} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{m \ne 2,m \ne 1}\\{m < \frac{4}{3},m > 2}\\{m < 1;m > 2}\\{1 < m < 2;m > 2}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{m < 1}\\{m > 2}\end{array}} \right.\).

Để f(x) = 0, f(x) có 2 nghiệm dương phân biệt

⇒ m ∈ (–∞; 1) ∪ (2; +∞).