Question

помогите решить дам пжл 35 балов

Answer 1

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

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