Answer:
[tex]\dfrac{c}{3(c+2)}[/tex]
Step-by-step explanation:
[tex]\dfrac{c^2+4c}{c^2-4} \div \dfrac{3c+12}{c-2}=\dfrac{c^2+4c}{c^2-4} \times \dfrac{c-2}{3c+12}=\dfrac{(c^2+4c)(c-2)}{(c^2-4)(3c+12)}[/tex]
Factor [tex]c^2-4[/tex]: [tex]c^2-4=(c-2)(c+2)[/tex]
[tex]\implies\dfrac{(c^2+4c)(c-2)}{(c+2)(c-2)(3c+12)}[/tex]
Cancel the common factor [tex](c-2)[/tex]:
[tex]\implies\dfrac{(c^2+4c)}{(c+2)(3c+12)}[/tex]
Factor [tex]c^2+4c[/tex]: [tex]c^2+4c=c(c+4)[/tex]
Factor [tex]3c+12[/tex]: [tex]3c+12=3(c+4)[/tex]
[tex]\implies\dfrac{c(c+4)}{3(c+2)(c+4)}[/tex]
Cancel the common factor [tex](c+4)[/tex]:
[tex]\implies\dfrac{c}{3(c+2)}[/tex]
