Answer:
[tex]\large\boxed{\dfrac{1}{(m-4)(m-3)}=\dfrac{1}{m^2-7m+12}}[/tex]
Step-by-step explanation:
[tex]\dfrac{\frac{m+3}{m^2-16}}{\frac{m^2-9}{m+4}}=\dfrac{m+3}{m^2-16}\cdot\dfrac{m+4}{m^2-9}=(*)\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\m^2-16=m^2=4^2=(m-4)(m+4)\\m^2-9=m^2-3^2=(m-3)(m+3)\\\\(*)=\dfrac{m+3}{(m-4)(m+4)}\cdot\dfrac{m+4}{(m-3)(m+3)}\\\\\text{cancel}\ (m+3)\ \text{and}\ (m+4)\\\\=\dfrac{1}{m-4}\cdot\dfrac{1}{m-3}=\dfrac{1}{(m-4)(m-3)}\\\\\text{use\ FOIL:}\ (a+b)(c+d)=ac+ad+bc+bd\\\\=\dfrac{1}{(m)(m)+(m)(-3)+(-4)(m)+(-4)(-3)}\\\\=\dfrac{1}{m^2-3m-4m+12}=\dfrac{1}{m^2-7m+12}[/tex]
