ĐK: $\dfrac{5.2^x-8}{2^x+2}>0$
$\iff 5.2^x-8>0$
....
$pt \iff \log _2 \Big(\dfrac{5.2^x-8}{2^x+2}\Big)=3-x$
$\iff \dfrac{5.2^x-8}{2^x+2}=2^{3-x}$
$\iff 5.2^x-8=\dfrac{2^3}{2^x} \cdot (2^x+2)$
$\iff 5.(2^x)^2-8.2^x=2^3.2^x+16$
$\iff 5.(2^x)^2-16.2^x-16=0$
$\iff \left[\begin{array}{l} 2^x=\dfrac{-4}{5} (l) \\2^x= 4 \end{array}\right.$
$\implies x=2$
Có gì không hiểu hỏi lại nha em