Google
Câu hỏi: Bất phương trình ${{\log }_{3}}{{\left( {{x}^{2}}+3 \right)}^{2019}}{{.2}^{{{x}^{2}}-12}}-{{2019}^{x+5}}.{{\log }_{\sqrt[16]{3}}}\left( {{x}^{2}}+3 \right)\le 0$ có tất cả bao nhiêu nghiệm nguyên?
A. 8.
B. 13.
C. 19.
D. 20.
A. 8.
B. 13.
C. 19.
D. 20.
${{\log }_{3}}{{\left( {{x}^{2}}+3 \right)}^{2019}}{{.2}^{{{x}^{2}}-12}}-{{2019}^{x+5}}.{{\log }_{\sqrt[16]{3}}}\left( {{x}^{2}}+3 \right)\le 0$
$\Leftrightarrow 2019.{{\log }_{3}}\left( {{x}^{2}}+3 \right){{.2}^{{{x}^{2}}-12}}-{{2019}^{x+5}}.16{{\log }_{3}}\left( {{x}^{2}}+3 \right)\le 0$
$\Leftrightarrow {{2019.2}^{{{x}^{2}}-12}}\le {{2019}^{x+5}}.16$ (do ${{\log }_{3}}\left( {{x}^{2}}+3 \right)>0,\forall x\in \mathbb{R}$ )
$\begin{aligned}
& \Leftrightarrow {{2}^{{{x}^{2}}-16}}\le {{2019}^{x+4}}\Leftrightarrow {{\log }_{2}}{{2}^{{{x}^{2}}-16}}\le {{\log }_{2}}{{2019}^{x+4}}\Leftrightarrow {{x}^{2}}-16\le \left( x+4 \right){{\log }_{2}}2019 \\
& \Leftrightarrow \left( x+4 \right)\left( x-4-{{\log }_{2}}2019 \right)\le 0\Leftrightarrow -4\le x\le 4+{{\log }_{2}}2019. \\
\end{aligned}$
Theo đề thì $x$ là nghiệm nguyên nên ta có: $x\in \left\{ -4,-3,-2,...,14 \right\}$
Vậy có 19 nghiệm nguyên.
$\Leftrightarrow 2019.{{\log }_{3}}\left( {{x}^{2}}+3 \right){{.2}^{{{x}^{2}}-12}}-{{2019}^{x+5}}.16{{\log }_{3}}\left( {{x}^{2}}+3 \right)\le 0$
$\Leftrightarrow {{2019.2}^{{{x}^{2}}-12}}\le {{2019}^{x+5}}.16$ (do ${{\log }_{3}}\left( {{x}^{2}}+3 \right)>0,\forall x\in \mathbb{R}$ )
$\begin{aligned}
& \Leftrightarrow {{2}^{{{x}^{2}}-16}}\le {{2019}^{x+4}}\Leftrightarrow {{\log }_{2}}{{2}^{{{x}^{2}}-16}}\le {{\log }_{2}}{{2019}^{x+4}}\Leftrightarrow {{x}^{2}}-16\le \left( x+4 \right){{\log }_{2}}2019 \\
& \Leftrightarrow \left( x+4 \right)\left( x-4-{{\log }_{2}}2019 \right)\le 0\Leftrightarrow -4\le x\le 4+{{\log }_{2}}2019. \\
\end{aligned}$
Theo đề thì $x$ là nghiệm nguyên nên ta có: $x\in \left\{ -4,-3,-2,...,14 \right\}$
Vậy có 19 nghiệm nguyên.
Đáp án C.