Đáp án:
\(\left[ \begin{array}{l}
x = 1\\
x = 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
{4^{\sqrt {x + 2} }}{.4^{2x}} + {2^{{x^3}}} = {4^{\sqrt {x + 2} }}{.4^2} + {2^{{x^3} + 4x - 4}}\\
\to {4^{\sqrt {x + 2} }}{.16^x} - {16.4^{\sqrt {x + 2} }} = {2^{{x^3}}}{.2^{4x - 4}} - {2^{{x^3}}}\\
\to {4^{\sqrt {x + 2} }}\left( {{{16}^x} - 16} \right) = {2^{{x^3}}}\left( {\dfrac{{{{16}^x}}}{{16}} - 1} \right)\\
\to {4^{\sqrt {x + 2} }}\left( {{{16}^x} - 16} \right) = {2^{{x^3}}}\left( {\dfrac{{{{16}^x} - 16}}{{16}}} \right)\\
\to {4^{\sqrt {x + 2} }}\left( {{{16}^x} - 16} \right) - \dfrac{{{2^{{x^3}}}}}{{16}}\left( {{{16}^x} - 16} \right) = 0\\
\to \left( {{{16}^x} - 16} \right)\left( {{4^{\sqrt {x + 2} }} - \dfrac{{{2^{{x^3}}}}}{{16}}} \right) = 0\\
\to \left[ \begin{array}{l}
{16^x} - 16 = 0\\
{4^{\sqrt {x + 2} }} - \dfrac{{{2^{{x^3}}}}}{{16}} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
{4^{{{\left( {x + 2} \right)}^{\dfrac{1}{2}}}}} - \dfrac{{{2^{{3^x}}}}}{{16}} = 0\left( * \right)
\end{array} \right.\\
\left( * \right) \to {4^{{{\dfrac{1}{2}}^{x + 2}}}} - \dfrac{{{8^x}}}{{16}} = 0\\
\to {2^{x + 2}} - \dfrac{{{8^x}}}{{16}} = 0\\
\to {4.2^x} = \dfrac{{{8^x}}}{{16}}\\
\to {64.2^x} = {2^{3x}}\\
\to {2^{3x}} - {64.2^x} = 0\\
\to {2^x}\left( {{2^{2x}} - 64} \right) = 0\\
\to {2^{2x}} - 64 = 0\\
\to {4^x} = 64\\
\to {4^x} = {4^3}\\
\to x = 3\\
KL:\left[ \begin{array}{l}
x = 1\\
x = 3
\end{array} \right.
\end{array}\)
