Question

алгебра срочно поомогите

Answer 1

$(\frac{m^{2}-3m+9}{9m^2-1}* \frac{3m^2+m}{m^3+27}-\frac{m+3}{3m^2-m}):\frac{7}{m^2+3m}-\frac{15m+6}{7-21m} \\1) \frac{m^{2}-3m+9}{9m^2-1}* \frac{3m^2+m}{m^3+27}=\frac{m^{2}-3m+9}{(3m-1)(3m+1)}*\frac{m(3m+1)}{(m+3)(m^2-3m+9)}=\frac{(m^{2}-3m+9)*m(3m+1)}{(3m-1)(3m+1)*(m+3)(m^2-3m+9)} =\frac{m}{(3m-1)(m+3)}$$2)\frac{m}{(3m-1)(m+3)}-\frac{m+3}{3m^2-m}= \frac{m}{(3m-1)(m+3)}-\frac{m+3}{m(3m-1)}=\frac{m*m}{(3m-1)(m+3)*m}-\frac{(m+3)*(m+3)}{m(3m-1)*(m+3)}=\frac{m*m-(m+3)*(m+3)}{m(3m-1)(m+3)} =\frac{m^2-(m^2+6m+9)}{m(3m-1)(m+3)}=\frac{m^2-m^2-6m-9}{m(3m-1)(m+3)}=\frac{-6m-9}{m(3m-1)(m+3)}$$3) \frac{-6m-9}{m(3m-1)(m+3)}:\frac{7}{m^2+3m} = \frac{-6m-9}{m(3m-1)(m+3)}*\frac{m(m+3)}{7} =\frac{(-6m-9)*m(m+3)}{7m(3m-1)(m+3)} =\frac{-6m-9}{7(3m-1)}$

$4)\frac{-6m-9}{7(3m-1)} -\frac{15m+6}{7-21m}=\frac{-6m-9}{7(3m-1)} -\frac{15m+6}{-7(3m-1)}=\frac{-6m-9}{7(3m-1)} -(-\frac{15m+6}{7(3m-1)})=\frac{-6m-9}{7(3m-1)} +\frac{15m+6}{7(3m-1)})=\frac{-6m-9+15m+6}{7(3m-1)} =\frac{9m-3}{7(3m-1)}=\frac{3(3m-1)}{7(3m-1)}=\frac{3}{7}$

Ответ: $\frac{3}{7}$