Question

(3/(9-m^2 )+(2m-1)/(m-3)-(m^2-4)/(m^2+6m+9)×(m+3)/(m-2))÷m/(m-3)

Answer 1

$\tt\displaystyle (\frac{3}{9-m^2} +\frac{2m-1}{m-3}-\frac{m^2-4}{m^2+6m+9}\cdot \frac{m+3}{m-2}): \frac{m}{m-3}=\\\\ (\frac{-3}{m^2-9}+\frac{2m-1}{m-3}-\frac{(m-2)(m+2)(m+3)}{(m+3)^2(m-2)})\cdot \frac{m-3}{m}=\\\\ (\frac{-3+(2m-1)(m+3)}{(m-3)(m+3)}-\frac{m+2}{m+3})\cdot \frac{m-3}{m}=\\\\ (\frac{2m^2+6m-m-3-3-(m^2+2m-3m-6)}{(m+3)(m-3)})\cdot \frac{m-3}{m}=\\\\ \frac{(m^2+6m)(m-3)}{(m+3)(m-3)m}=\frac{m+6}{m+3}$