Đáp án:
$\begin{array}{l}
{4^{{x^2} - 3x + 2}} + {4^{{x^2} + 6x + 5}} = {4^{\left( {{x^2} - 3x + 2} \right) + \left( {{x^2} + 6x + 5} \right)}} + 1\\
\Rightarrow {4^{{x^2} - 3x + 2}} + {4^{{x^2} + 6x + 5}} = {4^{{x^2} - 3x + 2}}{.4^{{x^2} + 6x + 5}} + 1\\
\Rightarrow {4^{{x^2} - 3x + 2}}\left( {1 - {4^{{x^2} + 6x + 5}}} \right) + \left( {{4^{{x^2} + 6x + 5}} - 1} \right) = 0\\
\Rightarrow \left( {1 - {4^{{x^2} + 6x + 5}}} \right).\left( {{4^{{x^2} - 3x + 2}} - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
{4^{{x^2} + 6x + 5}} = 1\\
{4^{{x^2} - 3x + 2}} = 1
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
{x^2} + 6x + 5 = 0\\
{x^2} - 3x + 2 = 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = - 1\\
x = - 5\\
x = 1\\
x = 2
\end{array} \right.
\end{array}$
