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Câu hỏi: Tập nghiệm của bất phương trình ${{\left( \dfrac{1}{2} \right)}^{9{{x}^{2}}-17x+11}}\ge {{\left( \dfrac{1}{2} \right)}^{7-5x}}$ là
A. $\left( \dfrac{2}{3};+\infty \right)$.
B. $\left\{ \dfrac{2}{3} \right\}$.
C. $\left( -\infty ;\dfrac{2}{3} \right)$.
D. $\mathbb{R}\backslash \left\{ \dfrac{2}{3} \right\}$.
A. $\left( \dfrac{2}{3};+\infty \right)$.
B. $\left\{ \dfrac{2}{3} \right\}$.
C. $\left( -\infty ;\dfrac{2}{3} \right)$.
D. $\mathbb{R}\backslash \left\{ \dfrac{2}{3} \right\}$.
Ta có: ${{\left( \dfrac{1}{2} \right)}^{9{{x}^{2}}-17x+11}}\ge {{\left( \dfrac{1}{2} \right)}^{7-5x}}\Leftrightarrow 9{{x}^{2}}-17x+11\le 7-5x\Leftrightarrow 9{{x}^{2}}-12x+4\le 0$
$\Leftrightarrow {{\left( 3x-2 \right)}^{2}}\le 0\Leftrightarrow x=\dfrac{2}{3}$.
$\Leftrightarrow {{\left( 3x-2 \right)}^{2}}\le 0\Leftrightarrow x=\dfrac{2}{3}$.
Đáp án B.