Cho phương trình 2x^2 + 4x − 5 = 0 có hai nghiệm là x1 , x2 .

Xét phương trình \(2{x^2} + 4x - 5 = 0\) có \(\Delta  = {4^2} - 4.2.\left( { - 5} \right) = 56 > 0\) nên phương trình có hai nghiệm phân biệt \({x_1}\); \({x_2}\).

                 Áp dụng định lí Viète, ta có: \(\left\{ {\begin{array}{*{20}{l}}{{x_1} + {x_2} =  - \frac{4}{2} =  - 2}\\{{x_1}{x_2} = \frac{{ - 5}}{2}}\end{array}} \right.\)

                   Ta có:           \(T = \frac{{2{x_1} - 1}}{{{x_2}}} + \frac{{2{x_2} - 1}}{{{x_1}}} + 2026\)

                                        \( = \frac{{{x_1}\left( {2{x_1} - 1} \right)}}{{{x_1}{x_2}}} + \frac{{{x_2}\left( {2{x_2} - 1} \right)}}{{{x_1}{x_2}}} + \frac{{2026{x_1}{x_2}}}{{{x_1}{x_2}}}\)

                                        \( = \frac{{2x_1^2 - {x_1} + 2x_2^2 - {x_2} + 2026{x_1}{x_2}}}{{{x_1}{x_2}}}\)

                                        \( = \frac{{\left( {2x_1^2 + 4{x_1}{x_2} + 2x_2^2} \right) - \left( {{x_1} + {x_2}} \right) + 2022{x_1}{x_2}}}{{{x_1}{x_2}}}\)

                                        \( = \frac{{2{{\left( {{x_1} + {x_2}} \right)}^2} - \left( {{x_1} + {x_2}} \right) + 2022{x_1}{x_2}}}{{{x_1}{x_2}}}\)

                                        \( = \frac{{2 \cdot {{( - 2)}^2} - \left( { - 2} \right) + 2022 \cdot \left( {\frac{{ - 5}}{2}} \right)}}{{\frac{{ - 5}}{2}}}\)

                                        \( = \frac{{8 + 2 - 5055}}{{\frac{{ - 5}}{2}}} = 2018\)

                     Vậy \(T = 2018\)