cho tam thức: f(x) = x² -2mx +4m -3 Tìm m để F(x) luôn dương với mọi x ∈ (0;1)

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1. Đáp án:

   \(m \ge 1\)

   Giải thích các bước giải:

   Ta có: \(\Delta ‘ = {m^2} – 4m + 3\)

   \(f\left( x \right)\) luôn dương với mọi \(x \in \left( {0;1} \right)\) \( \Leftrightarrow f\left( x \right) > 0,\forall x \in \mathbb{R}\) hoặc \(f\left( x \right) = 0\) có hai nghiệm \({x_1} \le {x_2}\) thỏa mãn \(\left[ \begin{array}{l}\left( {0;1} \right) \subset \left( { – \infty ;{x_1}} \right)\\\left( {0;1} \right) \subset \left( {{x_2}; + \infty } \right)\end{array} \right.\)

   TH1: \(f\left( x \right) > 0,\forall x \in \mathbb{R}\)\( \Leftrightarrow \left\{ \begin{array}{l}a = 1 > 0\\\Delta ‘ < 0\end{array} \right. \Leftrightarrow {m^2} – 4m + 3 < 0 \Leftrightarrow 1 < m < 3\)

   TH2: \(f\left( x \right) = 0\) có hai nghiệm \({x_1} \le {x_2}\) thỏa mãn \(\left[ \begin{array}{l}\left( {0;1} \right) \subset \left( { – \infty ;{x_1}} \right)\\\left( {0;1} \right) \subset \left( {{x_2}; + \infty } \right)\end{array} \right.\)

   \(\begin{array}{l}\left\{ \begin{array}{l}\Delta ‘ \ge 0\\\left[ \begin{array}{l}1 \le {x_1} \le {x_2}\\{x_1} \le {x_2} \le 0\end{array} \right.\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{m^2} – 4m + 3 \ge 0\\\left[ \begin{array}{l}\left\{ \begin{array}{l}\dfrac{S}{2} \ge 1\\a.f\left( 1 \right) \ge 0\end{array} \right.\\\left\{ \begin{array}{l}\dfrac{S}{2} \le 0\\a.f\left( 0 \right) \ge 0\end{array} \right.\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}m \ge 3\\m \le 1\end{array} \right.\\\left[ \begin{array}{l}\left\{ \begin{array}{l}m \ge 1\\1.\left( {{1^2} – 2m.1 + 4m – 3} \right) \ge 0\end{array} \right.\\\left\{ \begin{array}{l}m \le 0\\1.\left( {{0^2} – 2m.0 + 4m – 3} \right) \ge 0\end{array} \right.\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}m \ge 3,m \le 1\\\left[ \begin{array}{l}\left\{ \begin{array}{l}m \ge 1\\ – 2 + 2m \ge 0\end{array} \right.\\\left\{ \begin{array}{l}m \le 0\\4m – 3 \ge 0\end{array} \right.\end{array} \right.\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m \ge 3,m \le 1\\\left[ \begin{array}{l}m \ge 1\\\left\{ \begin{array}{l}m \le 0\\m \ge \dfrac{3}{4}\end{array} \right.\left( {VN} \right)\end{array} \right.\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m \ge 3,m \le 1\\m \ge 1\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}m \ge 3\\m = 1\end{array} \right.\end{array}\)

   Kết hợp với TH1 ta được \(\left[ \begin{array}{l}1 < m < 3\\m \ge 3\\m = 1\end{array} \right. \Leftrightarrow m \ge 1\).

   Vậy \(m \ge 1\).

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