[tex]\dfrac{3+4i}{2+4i}=\dfrac{3+4i}{2+4i}\cdot\dfrac{2-4i}{2-4i}=\dfrac{(3)(2)+(3)(-4i)+(4i)(2)+(4i)(-4i)}{2^2-(4i)^2}\\\\=\dfrac{6-12i+8i-16i^2}{4-16i^2}=\dfrac{6-4i-16(-1)}{4-16(-1)}=\dfrac{6-4i+16}{4+16}\\\\=\dfrac{22-4i}{20}=\boxed{\dfrac{11-2i}{10}}=\boxed{\dfrac{11}{10}-\dfrac{1}{5}i}[/tex]
