Giải thích các bước giải:
Ta có:
\[\begin{array}{l}
{10^x} + {2.5^{2x}} \ge {4^x}\\
\Leftrightarrow {10^x} + 2.{\left( {{5^2}} \right)^x} - {4^x} \ge 0\\
\Leftrightarrow {10^x} + {2.25^x} - {4^x} \ge 0\\
\Leftrightarrow \frac{{{{10}^x}}}{{{{25}^x}}} + 2 - \frac{{{4^x}}}{{{{25}^x}}} \ge 0\\
\Leftrightarrow {\left( {\frac{2}{5}} \right)^x} + 2 - {\left( {\frac{{{2^2}}}{{{5^2}}}} \right)^x} \ge 0\\
\Leftrightarrow {\left( {\frac{2}{5}} \right)^x} + 2 - {\left( {\frac{2}{5}} \right)^{2x}} \ge 0\\
\Leftrightarrow {\left( {\frac{2}{5}} \right)^{2x}} - {\left( {\frac{2}{5}} \right)^x} - 2 \le 0\\
\Leftrightarrow - 1 \le {\left( {\frac{2}{5}} \right)^x} \le 2\\
\Rightarrow x \ge {\log _{\frac{2}{5}}}2
\end{array}\]
