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Câu hỏi: Bất phương trình $\dfrac{{{2.3}^{x}}-{{2}^{x+2}}}{{{3}^{x}}-{{2}^{x}}}\le 1$ có bao nhiêu nghiệm nguyên?
A. 2.
B. 1.
C. 4.
D. vô số.
A. 2.
B. 1.
C. 4.
D. vô số.
$\dfrac{{{2.3}^{x}}-{{2}^{x+2}}}{{{3}^{x}}-{{2}^{x}}}\le 1\Leftrightarrow \dfrac{2.{{\left( \dfrac{3}{2} \right)}^{x}}-4}{{{\left( \dfrac{3}{2} \right)}^{x}}-1}\le 1\Leftrightarrow \dfrac{2.{{\left( \dfrac{3}{2} \right)}^{x}}-4}{{{\left( \dfrac{3}{2} \right)}^{x}}-1}-1\le 0$
$\Leftrightarrow \dfrac{{{\left( \dfrac{3}{2} \right)}^{x}}-3}{{{\left( \dfrac{3}{2} \right)}^{x}}-1}\le 0\Leftrightarrow 1<{{\left( \dfrac{3}{2} \right)}^{x}}\le 3\Leftrightarrow 0<x\le {{\log }_{\dfrac{3}{2}}}3.$
$\Rightarrow x\in \{1;2\}$ vậy có 2 nghiệm nguyên
$\Leftrightarrow \dfrac{{{\left( \dfrac{3}{2} \right)}^{x}}-3}{{{\left( \dfrac{3}{2} \right)}^{x}}-1}\le 0\Leftrightarrow 1<{{\left( \dfrac{3}{2} \right)}^{x}}\le 3\Leftrightarrow 0<x\le {{\log }_{\dfrac{3}{2}}}3.$
$\Rightarrow x\in \{1;2\}$ vậy có 2 nghiệm nguyên
Đáp án A.