Đáp án:
\(\begin{array}{l}
a)x = 4\\
c)\left[ \begin{array}{l}
x = 5\\
x = 6
\end{array} \right.\\
b)x \in \emptyset \\
d)\left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a){2^x} + {2^x}{.2^3} = 144\\
\to {2^x} + {8.2^x} = 144\\
\to {9.2^x} = 144\\
\to {2^x} = 16\\
\to {2^x} = {2^4}\\
\to x = 4\\
c){\left( {x - 5} \right)^4} = {\left( {x - 5} \right)^6}\\
\to {\left( {x - 5} \right)^6} - {\left( {x - 5} \right)^4} = 0\\
\to {\left( {x - 5} \right)^4}\left( {{{\left( {x - 5} \right)}^2} - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x - 5 = 0\\
{\left( {x - 5} \right)^2} = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 5\\
x - 5 = 1\\
x - 5 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 5\\
x = 6\\
x = 4\left( l \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 5\\
x = 6
\end{array} \right.\\
b){3^{2x}}{.3^2} = {9^x}{.9^3}\\
\to {9^x}.9 = {9^x}{.9^3}\\
\to {9^x}{.9^3} - {9^x}.9 = 0\\
\to {9^x}\left( {{9^3} - 9} \right) = 0\\
\to {720.9^x} = 0\\
\to {9^x} = 0\left( {KTM} \right)\\
Do:{9^x} > 0\forall x\\
\to x \in \emptyset \\
d){x^{15}} = {x^2}\\
\to {x^{15}} - {x^2} = 0\\
\to {x^2}\left( {{x^{13}} - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
{x^{13}} - 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.
\end{array}\)
