Помогите ребята. 1)(2-m/2+m-m+2/m-2)÷(2+m/2-m+m-2/m+2)

Решите задачу:

$( \frac{2-m}{2+m}- \frac{m+2}{m-2}) :( \frac{2+m}{2-m}+ \frac{m-2}{m+2})= (\frac{2-m}{2+m}+ \frac{m+2}{2-m}):( \frac{2+m}{2-m}- \frac{2-m}{m+2})= \\ \\ \frac{(2-m)(2-m)+(m+2)(m+2)}{(2+m)(2-m)}* \frac{(2-m)(2+m)}{(2+m)(2+m)-(2-m)(2-m)}= \frac{(2-m)^{2}+(m+2)^{2} }{(2+m)^{2} -(2-m)^{2} } = \\ \\ \frac{4-4m+m^{2}+m^{2}+4m+4 }{(2+m-2+m)(2+m+2-m)}= \frac{8+2m^{2} }{8m}= \frac{2(4+m^{2} )}{8m}= \frac{4+ m^{2} }{4m}$