A.\(m \ge 0\)B.\(m < - \dfrac{5}{3}\)C.\(m < 1\)D.\(

lop92lua

2 năm trước

A.\(m \ge 0\)
B.\(m < - \dfrac{5}{3}\)
C.\(m < 1\)
D.\( - \dfrac{5}{3} \le m < 1\)

Loga Hóa Học lớp 12

0 lượt thích 22 xem

1 trả lời

Thích Trả lời

led586490

Đáp án đúng: D

Phương pháp giải:

Áp dụng lý thuyết:\(f\left( x \right) = a{x^2} + bx + c\,\, \ge 0\), \(\forall x \in \mathbb{R} \Leftrightarrow \left\{ \begin{array}{l}a > 0\\\Delta \le 0\end{array} \right.\).\(f\left( x \right) = a{x^2} + bx + c > 0\), \(\forall x \in \mathbb{R} \Leftrightarrow \left\{ \begin{array}{l}a > 0\\\Delta < 0\end{array} \right.\).Giải chi tiết:Ta có:\(\begin{array}{l} - 1 \le \dfrac{{{x^2} + 5x + m}}{{2{x^2} - 3x + 2}} < 7\\ \Leftrightarrow \left\{ \begin{array}{l} - 1 \le \dfrac{{{x^2} + 5x + m}}{{2{x^2} - 3x + 2}}\\\dfrac{{{x^2} + 5x + m}}{{2{x^2} - 3x + 2}} < 7\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\dfrac{{ - 2{x^2} + 3x - 2}}{{2{x^2} - 3x + 2}} - \dfrac{{{x^2} + 5x + m}}{{2{x^2} - 3x + 2}} \le 0\\\dfrac{{{x^2} + 5x + m}}{{2{x^2} - 3x + 2}} - \dfrac{{14{x^2} - 21x + 14}}{{2{x^2} - 3x + 2}} < 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\dfrac{{ - 3{x^2} - 2x - m - 2}}{{2{x^2} - 3x + 2}} \le 0\\\dfrac{{ - 13{x^2} + 26x + m - 14}}{{2{x^2} - 3x + 2}} < 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} - 3{x^2} - 2x - m - 2 \le 0\\ - 13{x^2} + 26x + m - 14 < 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}3{x^2} + 2x + m + 2 \ge 0\,\,\,\,\left( 1 \right)\\13{x^2} - 26x - m + 14 > 0\,\,\,\,\left( 2 \right)\end{array} \right.\end{array}\)*) Xét \(\left( 1 \right)\):\(3{x^2} + 2x + m + 2 \ge 0,\,\,\forall x \in \mathbb{R}\)\( \Leftrightarrow \left\{ \begin{array}{l}a = 3 > 0\\\Delta \le 0\end{array} \right.\)\( \Leftrightarrow {2^2} - 4.3.\left( {m + 2} \right) \le 0\)\( \Leftrightarrow m \ge - \dfrac{5}{3}\)*) Xét \(\left( 2 \right)\):\(13{x^2} - 26x - m + 14 > 0,\,\,\forall x \in \mathbb{R}\)\( \Leftrightarrow \left\{ \begin{array}{l}a = 13 > 0\\\Delta < 0\end{array} \right.\)\( \Leftrightarrow {13^2} - 13.\left( { - m + 14} \right) \le 0\)\( \Leftrightarrow m < 1\)Vậy \( - \dfrac{5}{3} \le m < 1\).Đáp án D.

Vote (0) Phản hồi (0) 2 năm trước

Các câu hỏi liên quan

A.\(2\)
B.\(3\)
C.\(4\)
D.\(5\)

A.\(m \ge - \dfrac{5}{8}\)
B.\(m < \dfrac{5}{8}\)
C.\(m < - \dfrac{5}{8}\)
D.\(m \ge \dfrac{5}{8}\)

A.\(x \in \left( { - 4;\,\, - 3} \right) \cup \left( {2;\,\, + \infty } \right)\)
B.\(x \in \left( { - 4;\,\, - 3} \right) \cup \left[ {2;\,\, + \infty } \right)\)
C.\(x \in \left( { - \infty ;\,\, - 3} \right] \cup \left[ {2;\,\, + \infty } \right)\)
D.\(x \in \left( { - 4;\,\, - 3} \right] \cup \left[ {2;\,\, + \infty } \right)\)

A.\(S = \left( { - \infty ;\,\,1} \right) \cup \left( {4;\,\, + \infty } \right)\)
B.\(S = \left( { - \infty ;\,\,2} \right) \cup \left( {4;\,\, + \infty } \right)\)
C.\(S = \left( { - \infty ;\,\,2} \right) \cup \left( {3;\,\, + \infty } \right)\)
D.\(S = \left( { - \infty ;\,\,1} \right) \cup \left( {3;\,\, + \infty } \right)\)

A.\( - 14 \le m \le 2\)
B.\(m \le - 14\) hoặc \(m \ge 2\)
C.\(m < - 14\) hoặc \(m > 2\)
D.\( - 14 < m < 2\)

A.\(m > 28\)
B.\(m < 1\)
C.\(0 \le m \le 28\)
D.\(0 < m < 28\)

A.\(m \ge - 1\)
B.\( - 1 \le m \le 0\)
C.\(m \ge 0\)
D.\(m > - 1\)

A.\(\left( { - \infty ;\,\,\dfrac{1}{2}} \right)\)
B.\(\left( { - \infty ;\,\,\dfrac{1}{2}} \right]\)
C.\(\left( {\dfrac{1}{2};\,\, + \infty } \right)\)
D.\(\left[ {\dfrac{1}{2};\,\, + \infty } \right)\)

A.\(\left\{ \begin{array}{l}{x^2} - 2x - 3 > 0\\ - 2{x^2} + x - 1 < 0\end{array} \right.\)
B.\(\left\{ \begin{array}{l}{x^2} - 2x - 3 < 0\\ - 2{x^2} + x - 1 > 0\end{array} \right.\)
C.\(\left\{ \begin{array}{l}{x^2} - 2x - 3 > 0\\2{x^2} + x + 1 > 0\end{array} \right.\)
D.\(\left\{ \begin{array}{l}{x^2} - 2x - 3 < 0\\2{x^2} - x + 1 > 0\end{array} \right.\)

A.\( - \dfrac{{20}}{9} \le m \le 0\)
B.\(m \le 0\)
C.\(m \ge - \dfrac{{20}}{9}\)
D.\(m > 0\)

Loga.vn - Cộng Đồng Luyện Thi Trực Tuyến