Đáp án:
\(x > {\log _{\dfrac{5}{2}}}\dfrac{1}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
{2^{2x}} - {2.5^{2x}} < {2^x}{.5^x}\\
\to {2^{2x}} - {2.5^{2x}} - {2^x}{.5^x} < 0\\
\to \dfrac{{{2^{2x}}}}{{{2^{2x}}}} - \dfrac{{{{2.5}^{2x}}}}{{{2^{2x}}}} - \dfrac{{{5^x}}}{{{2^x}}} < 0\\
\to - 2.{\left( {\dfrac{5}{2}} \right)^{2x}} - \dfrac{{{5^x}}}{{{2^x}}} + 1 < 0\\
\to \left[ \begin{array}{l}
\dfrac{{{5^x}}}{{{2^x}}} > \dfrac{1}{2}\\
\dfrac{{{5^x}}}{{{2^x}}} < - 1\left( l \right)
\end{array} \right.\\
\to {\left( {\dfrac{5}{2}} \right)^x} > \dfrac{1}{2}\\
\to x > {\log _{\dfrac{5}{2}}}\dfrac{1}{2}
\end{array}\)
