Đáp án:
a) \(\left[ \begin{array}{l}
x = 1\\
x = 5
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:m = 2\\
Pt \to {x^2} - 6x + 5 = 0\\
\to \left( {x - 1} \right)\left( {x - 5} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1\\
x = 5
\end{array} \right.\\
b)Xét:\Delta ' > 0\\
\to {m^2} + 2m + 1 - 2m - 1 > 0\\
\to {m^2} > 0\\
\to m \ne 0\\
{x_1}^3 + {x_2}^3 = 2019\\
\to \left( {{x_1} + {x_2}} \right)\left( {{x_1}^2 - {x_1}{x_2} + {x_2}^2} \right) = 2019\\
\to \left( {{x_1} + {x_2}} \right)\left[ {\left( {{x_1}^2 + 2{x_1}{x_2} + {x_2}^2} \right) - 3{x_1}{x_2}} \right] = 2019\\
\to \left( {{x_1} + {x_2}} \right)\left[ {{{\left( {{x_1} + {x_2}} \right)}^2} - 3{x_1}{x_2}} \right] = 2019\\
\to \left( {2m + 2} \right)\left( {4{m^2} + 8m + 4 - 3\left( {2m + 1} \right)} \right) = 2019\\
\to \left( {2m + 2} \right)\left( {4{m^2} + 8m + 4 - 6m - 3} \right) = 2019\\
\to \left( {2m + 2} \right)\left( {4{m^2} + 2m + 1} \right) = 2019\\
\to 8{m^3} + 8{m^2} + 4{m^2} + 4m + 2m + 2 = 2019\\
\to 8{m^3} + 12{m^2} + 6m - 2017 = 0\\
\to m = 5,818447651
\end{array}\)
