Đáp án:
d) x=0
Giải thích các bước giải:
\(\begin{array}{l}
b){2.2^x} + {2^{ - 1}}{.2^x} + {2^x} = 28\\
\to \left( {2 + \dfrac{1}{2} + 1} \right){2^x} = 28\\
\to \dfrac{7}{2}{.2^x} = 28\\
\to {2^x} = 8\\
\to {2^x} = {2^3}\\
\to x = 3\\
d){3.2^{2x}} - {2.2^x}{.3^x} = {3^{2x}}\\
\to {3.2^{2x}} - {2.2^x}{.3^x} - {3^{2x}} = 0\\
\to 3 - 2{\left( {\dfrac{3}{2}} \right)^x} - {\left( {\dfrac{3}{2}} \right)^{2x}} = 0\\
\to \left[ \begin{array}{l}
{\left( {\dfrac{3}{2}} \right)^x} = 1\\
{\left( {\dfrac{3}{2}} \right)^x} = - 3\left( l \right)
\end{array} \right.\\
\to x = 0
\end{array}\)
