$\Leftrightarrow 2^{\frac{3x}{x+2}} = 2^2 . 3^2 .3^{2-x}$
$\Leftrightarrow 2^{\frac{3x}{x+2}-2} =3^{4-x}$
$\Leftrightarrow 2^{\frac{x-4}{x+2}} =3^{4-x}$
$\Leftrightarrow \frac{x-4}{x+2}= (4-x)\log_2 3$
$\Leftrightarrow x-4 =(4-x)(x+2) \log_2 3$
+ $x=4$ là 1 nghiệm
+ $(x+2) \log_2 3=-1$ tự giải nốt