nhaaaaaaaaaaaaaaaaaa Question nhaaaaaaaaaaaaaaaaaa in progress 0 Môn Toán Diễm Kiều 4 years 2021-04-29T10:54:32+00:00 2021-04-29T10:54:32+00:00 1 Answers 16 views 0
Answers ( )
Đáp án:
\(0 < m < \dfrac{{99}}{{20}}\)
Giải thích các bước giải:
Để phương trình có nghiệm
\(\begin{array}{l}
\to {m^2} – 4\left( {m – 4} \right) \ge 0\\
\to {m^2} – 4m + 16 \ge 0\\
\to {\left( {m – 2} \right)^2} + 12 \ge 0\left( {ld} \right)\forall m\\
Có:\left( {5{x_1} – 1} \right)\left( {5{x_2} – 1} \right) < 0\\
\to 25{x_1}{x_2} – 5\left( {{x_1} + {x_2}} \right) + 1 < 0\\
\to 25\left( {m – 4} \right) – 5m + 1 < 0\\
\to 25m – 100 – 5m + 1 < 0\\
\to 20m < 99\\
\to m < \dfrac{{99}}{{20}}\\
KL:0 < m < \dfrac{{99}}{{20}}
\end{array}\)