Resposta :
[tex]\Large\boxed{\begin{array}{l}\sf\dfrac{6}{x+2}=\dfrac{4}{2x}\\\sf6\cdot2x=4\cdot(x+2)\\\sf12x=4x+8\\\sf12x-4x=8\\\sf 8x=8\\\sf x=\dfrac{8}{8}\\\\\huge\boxed{\boxed{\boxed{\boxed{\sf x=1}}}} \end{array}}[/tex]
[tex]\Large\boxed{\begin{array}{l}\sf\dfrac{6}{x+2}=\dfrac{4}{2x}\\\sf6\cdot2x=4\cdot(x+2)\\\sf12x=4x+8\\\sf12x-4x=8\\\sf 8x=8\\\sf x=\dfrac{8}{8}\\\\\huge\boxed{\boxed{\boxed{\boxed{\sf x=1}}}} \end{array}}[/tex]