Question

Помогите,чет туплю)
$\frac{m}{ m^{2} -4} : ( \frac{4+ m^{2} }{2+m} -m)=$

Answer 1

m/(m^2-4):((4+m^2)/(2+m) - m) = m/((m-2)(m+2)):(4+m^2-m(m+2))/(m+2) =
= m/((m-2)(m+2))*(m+2)/(4+m^2-m^2-2m) = m/(m-2)*1/(4-2m) = -m/(2(m-2)^2)

Answer 2

$\frac{m}{m^{2}-4}:( \frac{4+m^{2}}{2+m} -m )= \frac{m}{m^{2}-4}:( \frac{4+m^{2}}{2+m} - \frac{m(2+m)}{2+m} )=\frac{m}{m^{2}-4}:( \frac{4+m^{2}-m(2+m}{2+m})=\frac{m}{m^{2}-4}:( \frac{4+m^{2}-2m+m^{2}}{2+m})= \frac{m}{m^{2}-4}:( \frac{4-2m}{2+m})=\frac{m}{-(4-m^{2})}:( \frac{2(2-m)}{2+m})=\frac{m}{-(4-m^{2})}*( \frac{2+m)}{2(2-m)})= \frac{m(2+m)}{-(4-m^{2})(2(2-m))}=\frac{m(2+m)}{-(2-m)(2+m)(2(2-m))} = \frac{m}{-2(2-m)(2-m)}= - \frac{m}{2(2-m)^{2}}$