Question

Пожалуйста помогите [(9k^2-m^2)/k] * [(k/(3k+m) - k/m)]

Answer 1

$\frac{9 k^{2} - m^{2} }{k}$ ·$( \frac{k}{3k+m} - \frac{k}{m} )$=
9k²-m² · $( \frac{1}{3k+m} - \frac{1}{m} )$=$\frac{(3k+m)(3k-m)(m-(3k+m))}{m(3k+m)}$ = $\frac{-3k(3k-m)}{m}$