$8^x+4=4^x+_{}$ $2^{x+2}$
$⇔4^x.2^x+4-4^x-4.2^x=0_{}$
$⇔4^x.(2^x-1)-4.(2^x-1)=0_{}$
$⇔(4^x-4)(2^x-1)=0_{}$
$⇔_{}$ \(\left[ \begin{array}{l}4^x-4=0\\2^x-1=0\end{array} \right.\) $⇔_{}$ \(\left[ \begin{array}{l}x=1\\x=0\end{array} \right.\)
$Vậy_{}$ $x_{1}=1;$ $x_{2}=0$