решите во ложениях там обведено

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

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

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

$=(\frac{m^2+2m}{2-m})\cdot(\frac{m-2}{m^2+2m})=\frac{m-2}{2-m}=-1$

$\frac{x^3y-xy^3}{xy^2-x2y}=\frac{xy(x^2-y^2)}{xy(y-1)}=\frac{(x-y)(x+y)}{y-1}$

$\frac{(1/3-1/4)(1/3+1/4)}{1/4-1}=\frac{(4/12-3/12)(4/12+3/12)}{1/4-4/4}=\frac{1/12\cdot7/12}{-3/4}=\frac{7}{-18}=-\frac{7}{18}$