Đáp án:
\(\left[ \begin{array}{l}
m = 0\\
m = 2022
\end{array} \right.\)
Giải thích các bước giải:
Để phương trình \({x^2} - mx + m - 1 = 0\left( 1 \right)\) có nghiệm
\(\begin{array}{l}
\to {m^2} - 4\left( {m - 1} \right) \ge 0\\
\to {m^2} - 4m + 4 \ge 0\\
\to {\left( {m - 2} \right)^2} \ge 0\left( {ld} \right)\forall m\\
\dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} = \dfrac{{{x_1} + {x_2}}}{{2021}}\\
\to \dfrac{{{x_1} + {x_2}}}{{{x_1}{x_1}}} = \dfrac{{{x_1} + {x_2}}}{{2021}}\\
\to \left( {{x_1} + {x_2}} \right)\left( {\dfrac{1}{{{x_1}{x_1}}} - \dfrac{1}{{2021}}} \right) = 0\\
\to \left[ \begin{array}{l}
{x_1} + {x_2} = 0\\
\dfrac{1}{{{x_1}{x_1}}} - \dfrac{1}{{2021}} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = 0\\
\dfrac{1}{{m - 1}} - \dfrac{1}{{2021}} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = 0\\
m - 1 = 2021
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = 0\\
m = 2022
\end{array} \right.
\end{array}\)
