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

Question

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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2 tháng 2021-07-24T19:18:19+00:00 1 Answers 3 views 0

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 $$m \ge 1$$.