Đáp án:

m=1

Giải thích các bước giải:

Để pt có 2 nghiệm phân biệt

⇔Δ>0

\(\begin{array}{l}
{m^2} - 2m + 1 - 4\left( {5m - 6} \right) > 0\\
 \to {m^2} - 2m + 1 - 20m + 24 > 0\\
 \to {m^2} - 22m + 25 > 0\\
 \to \left[ \begin{array}{l}
m > 11 + 4\sqrt 6 \\
m < 11 - 4\sqrt 6 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \frac{{1 - m + \sqrt {{m^2} - 22m + 25} }}{2}\\
x = \frac{{1 - m - \sqrt {{m^2} - 22m + 25} }}{2}
\end{array} \right.\\
Có:4{x_1} + 3{x_2} = 1\\
 \to 3\left( {{x_1} + {x_2}} \right) + {x_1} = 1\\
 \to \left[ \begin{array}{l}
3.\left( {1 - m} \right) + \frac{{1 - m + \sqrt {{m^2} - 22m + 25} }}{2} = 1\\
3.\left( {1 - m} \right) + \frac{{1 - m - \sqrt {{m^2} - 22m + 25} }}{2} = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
3 - 3m + \frac{{1 - m + \sqrt {{m^2} - 22m + 25} }}{2} = 1\\
3 - 3m + \frac{{1 - m - \sqrt {{m^2} - 22m + 25} }}{2} = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
1 - m + \sqrt {{m^2} - 22m + 25}  = 2\left( {3m - 2} \right)\\
1 - m - \sqrt {{m^2} - 22m + 25}  = 2\left( {3m - 2} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\sqrt {{m^2} - 22m + 25}  = 6m - 4 - 1 + m\\
\sqrt {{m^2} - 22m + 25}  = 1 - m - 6m + 4
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\sqrt {{m^2} - 22m + 25}  = 7m - 5\\
\sqrt {{m^2} - 22m + 25}  =  - 7m + 5
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m \ge \frac{5}{7}\\
{m^2} - 22m + 25 = {\left( {7m - 5} \right)^2}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m \ge \frac{5}{7}\\
{m^2} - 22m + 25 = 49{m^2} - 70m + 25
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m \ge \frac{5}{7}\\
48{m^2} - 48m = 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m \ge \frac{5}{7}\\
\left[ \begin{array}{l}
m = 0\left( l \right)\\
m = 1\left( {TM} \right)
\end{array} \right.
\end{array} \right.
\end{array}\)