Đáp án:
b) \(\left[ \begin{array}{l}
m = 3\\
m = - \frac{3}{4}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:m = - 3\\
Pt \to {x^2} - 6x + 5 = 0\\
\to \left( {x - 1} \right)\left( {x - 5} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1\\
x = 5
\end{array} \right.\\
b)Xét:\Delta ' > 0\\
\to {m^2} - 2m + 1 > 0\\
\to {\left( {m - 1} \right)^2} > 0\\
\to m \ne 1\\
Có:2\left( {{x_1}^2 + {x_2}^2} \right) - 5{x_1}{x_2} = 27\\
\to 2\left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2}} \right) - 9{x_1}{x_2} = 27\\
\to 2{\left( {{x_1} + {x_2}} \right)^2} - 9{x_1}{x_2} = 27\\
\to 2.4{m^2} - 9\left( {2m - 1} \right) = 27\\
\to 8{m^2} - 18m + 9 = 27\\
\to 8{m^2} - 18m - 18 = 0\\
\end{array}\)
\( \to \left[ \begin{array}{l}
m = 3\\
m = - \frac{3}{4}
\end{array} \right.\)
