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Câu hỏi: Giải bất phương trình ${{\left( \dfrac{3}{4} \right)}^{2x-4}}>{{\left( \dfrac{3}{4} \right)}^{x+1}}.$
A. $S=\left[ 5;+\infty \right)$
B. $S=\left( -\infty ;5 \right)$
C. $S=\left( -\infty ;-1 \right)$
D. $S=\left( -1;2 \right)$
A. $S=\left[ 5;+\infty \right)$
B. $S=\left( -\infty ;5 \right)$
C. $S=\left( -\infty ;-1 \right)$
D. $S=\left( -1;2 \right)$
Vì $\dfrac{3}{4}<1$ khi đó ${{\left( \dfrac{3}{4} \right)}^{2x-4}}>{{\left( \dfrac{3}{4} \right)}^{x+1}}\Rightarrow 2x-4<x+1\Leftrightarrow x<5$
Vậy $S=\left( -\infty ;5 \right).$
Vậy $S=\left( -\infty ;5 \right).$
Đáp án B.