Đáp án:
\(\left[ \begin{array}{l}
m = 2\\
m = \frac{1}{2}
\end{array} \right.\)
Giải thích các bước giải:
Để phương trình có nghiệm kép
⇔ Δ'=0
\(\begin{array}{l}
\to \left\{ \begin{array}{l}
4{m^2} + 20m + 25 - \left( {2m - 7} \right)\left( { - 14m + 1} \right) = 0\\
m \ne \frac{7}{2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne \frac{7}{2}\\
4{m^2} + 20m + 25 + 28{m^2} - 100m + 7 = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne \frac{7}{2}\\
32{m^2} - 80m + 32 = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne \frac{7}{2}\\
16\left( {m - 2} \right)\left( {2m - 1} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = 2\\
m = \frac{1}{2}
\end{array} \right.\left( {TM} \right)\\
\to \left[ \begin{array}{l}
x = \frac{{ - 2m - 5}}{{2m - 7}} = \frac{{ - 2.2 - 5}}{{2.2 - 7}} = 3\\
x = \frac{{ - 2m - 5}}{{2m - 7}} = \frac{{ - 2.\frac{1}{2} - 5}}{{2.\frac{1}{2} - 7}} = 1
\end{array} \right.
\end{array}\)
