Question

A : 6x²+7x-3>0

B : 6x² + 7x - 3 < 0

C : 3-2x-x²>0

D : 3-2x-x²<0

Answer 1

\( a)6{x^2} + 7x - 3 < 0\\ \Leftrightarrow 6{x^2} + 9x - 2x - 3 < 0\\ \Leftrightarrow 3x\left( {2x + 3} \right) - \left( {2x + 3} \right) < 0\\ \Leftrightarrow \left( {2x + 3} \right)\left( {3x - 1} \right) < 0\\ \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 2x + 3 < 0\\ 3x - 1 > 0 \end{array} \right.\\ \left\{ \begin{array}{l} 2x + 3 > 0\\ 3x - 1 < 0 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x < - \dfrac{3}{2}\\ x > \dfrac{1}{3} \end{array} \right.\\ \left\{ \begin{array}{l} x < - \dfrac{3}{2}\\ x < \dfrac{1}{3} \end{array} \right. \end{array} \right. \Leftrightarrow x \in \left( { - \dfrac{3}{2};\dfrac{1}{3}} \right) \)

Answer 2

Answer 3

(Câu trả lời bằng hình ảnh)

Answer 4

\( a)6{x^2} + 7x - 3 > 0\\ \Leftrightarrow 6{x^2} + 9x - 2x - 3 > 0\\ \Leftrightarrow 3x\left( {2x + 3} \right) - \left( {2x + 3} \right) > 0\\ \Leftrightarrow \left( {2x + 3} \right)\left( {3x - 1} \right) > 0\\ \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 2x + 3 > 0\\ x - 1 > 0 \end{array} \right.\\ \left\{ \begin{array}{l} 2x + 3 < 0\\ x - 1 < 0 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x > - \dfrac{3}{2}\\ x > 1 \end{array} \right.\\ \left\{ \begin{array}{l} x < - \dfrac{3}{2}\\ x < 1 \end{array} \right. \end{array} \right. \Leftrightarrow x \in \left( { - \infty ; - \dfrac{3}{2}} \right) \cup \left( {\dfrac{1}{3}; + \infty } \right) \)

Answer 5

Nhầm :))

\( a)6{x^2} + 7x - 3 > 0\\ \Leftrightarrow 6{x^2} + 9x - 2x - 3 > 0\\ \Leftrightarrow 3x\left( {2x + 3} \right) - \left( {2x + 3} \right) > 0\\ \Leftrightarrow \left( {2x + 3} \right)\left( {3x - 1} \right) > 0\\ \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 2x + 3 > 0\\ 3x - 1 > 0 \end{array} \right.\\ \left\{ \begin{array}{l} 2x + 3 < 0\\ 3x - 1 < 0 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x > - \dfrac{3}{2}\\ x > \dfrac{1}{3} \end{array} \right.\\ \left\{ \begin{array}{l} x < - \dfrac{3}{2}\\ x < \dfrac{1}{3} \end{array} \right. \end{array} \right. \Leftrightarrow x \in \left( { - \infty ; - \dfrac{3}{2}} \right) \cup \left( {\dfrac{1}{3}; + \infty } \right) \)