Đáp án:
\(\left[ \begin{array}{l}
m = - 12\\
m = \dfrac{{12}}{{13}}
\end{array} \right.\)
Giải thích các bước giải:
Để phương trình có 2 nghiệm $x_1,x_2$
⇒Δ'≥0
\(\begin{array}{l}
\to {m^2} - 4m + 4 ≥ 0\\
\to {\left( {m - 2} \right)^2} ≥ 0(ld)\\
\to \left[ \begin{array}{l}
x = m + \sqrt {{{\left( {m - 2} \right)}^2}} \\
x = m - \sqrt {{{\left( {m - 2} \right)}^2}}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = m + m - 2\\
x = m - m + 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2m - 2\\
x = 2
\end{array} \right.\\
Có:3{x_1} + 5{x_2} = \dfrac{{17m}}{3}\\
\to \left[ \begin{array}{l}
3\left( {2m - 2} \right) + 5.2 = \dfrac{{17m}}{3}\\
3.2 + 5\left( {2m - 2} \right) = \dfrac{{17m}}{3}
\end{array} \right.\\
\to \left[ \begin{array}{l}
6m - 6 + 10 = \dfrac{{17m}}{3}\\
6 + 10m - 10 = \dfrac{{17m}}{3}
\end{array} \right.\\
\to \left[ \begin{array}{l}
\dfrac{1}{3}m = - 4\\
\dfrac{{13}}{3}m = 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = - 12\\
m = \dfrac{{12}}{{13}}
\end{array} \right.
\end{array}\)
