TRẢ LỜI

1. Giải đáp:

   $\begin{array}{l}
   {x^2} – 6x + m + 4 = 0\\
   a)Khi:m = 1\\
    \Leftrightarrow {x^2} – 6x + 5 = 0\\
    \Leftrightarrow \left( {x – 1} \right)\left( {x – 5} \right) = 0\\
    \Leftrightarrow x = 1;x = 5\\
   Vậy\,x = 1;x = 5\,khi:m = 1\\
   b)\Delta ‘ > 0\\
    \Leftrightarrow 9 – m – 4 > 0\\
    \Leftrightarrow m < 5\\
   Theo\,Viet:\left\{ \begin{array}{l}
   {x_1} + {x_2} = 6\\
   {x_1}{x_2} = m + 4
   \end{array} \right.\\
   2020.\left( {{x_1} + {x_2}} \right) – 2021.{x_1}{x_2} = 2014\\
    \Leftrightarrow 2020.6 – 2021.\left( {m + 4} \right) = 2014\\
    \Leftrightarrow m + 4 = \dfrac{{10106}}{{2021}}\\
    \Leftrightarrow m = \dfrac{{2022}}{{2021}}\left( {tmdk} \right)\\
   Vậy\,m = \dfrac{{2020}}{{2021}}
   \end{array}$

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