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Câu hỏi: Bất phương trình ${{\left( 0,2 \right)}^{{{x}^{2}}}}{{.2}^{x}}\ge \dfrac{2}{5}$ tương đương với bất phương trình nào sau đây?
A. $-{{x}^{2}}+x-{{\log }_{2}}\left( \dfrac{2}{5} \right)\ge 0.$
B. ${{x}^{2}}-x{{\log }_{5}}2+{{\log }_{5}}2-1\ge 0.$
C. $x\ge 1.$
D. ${{x}^{2}}-x{{\log }_{5}}2+{{\log }_{5}}2-1\le 0.$
A. $-{{x}^{2}}+x-{{\log }_{2}}\left( \dfrac{2}{5} \right)\ge 0.$
B. ${{x}^{2}}-x{{\log }_{5}}2+{{\log }_{5}}2-1\ge 0.$
C. $x\ge 1.$
D. ${{x}^{2}}-x{{\log }_{5}}2+{{\log }_{5}}2-1\le 0.$
Ta có:
$\begin{aligned}
& {{\left( 0,2 \right)}^{{{x}^{2}}}}{{.2}^{x}}\ge \dfrac{2}{5}\Leftrightarrow {{\log }_{5}}\left[ {{\left( 0,2 \right)}^{{{x}^{2}}}}{{.2}^{x}} \right]\ge {{\log }_{5}}\dfrac{2}{5}\Leftrightarrow {{x}^{2}}{{\log }_{5}}\left( 0,2 \right)+x{{\log }_{5}}2\ge {{\log }_{5}}2-{{\log }_{5}}5 \\
& \Leftrightarrow -{{x}^{2}}+x{{\log }_{5}}2-{{\log }_{5}}2+1\ge 0\Leftrightarrow {{x}^{2}}-x{{\log }_{5}}2+{{\log }_{5}}2-1\le 0 \\
\end{aligned}$
$\begin{aligned}
& {{\left( 0,2 \right)}^{{{x}^{2}}}}{{.2}^{x}}\ge \dfrac{2}{5}\Leftrightarrow {{\log }_{5}}\left[ {{\left( 0,2 \right)}^{{{x}^{2}}}}{{.2}^{x}} \right]\ge {{\log }_{5}}\dfrac{2}{5}\Leftrightarrow {{x}^{2}}{{\log }_{5}}\left( 0,2 \right)+x{{\log }_{5}}2\ge {{\log }_{5}}2-{{\log }_{5}}5 \\
& \Leftrightarrow -{{x}^{2}}+x{{\log }_{5}}2-{{\log }_{5}}2+1\ge 0\Leftrightarrow {{x}^{2}}-x{{\log }_{5}}2+{{\log }_{5}}2-1\le 0 \\
\end{aligned}$
Đáp án D.