Question

m^(-6)-27*27 *m^2/(9-6/m+1/m^2)(9+3/m+1/m^2)+ 9m^2(3m+1)/ (3m-1)
Ответ:3m+1

Answer 1

$\frac{( m^{-6} -27*27)*m^2}{(9- \frac{6}{m}+ \frac{1}{m^2} )(9+ \frac{3}{m} + \frac{1}{m^2} ) } + \frac{9m^2(1+3m)}{3m-1} = \\ \\ \frac{ \frac{1}{m^4}-(27m)^2 }{ \frac{9m^2-6m+1}{m^2}* \frac{9m^2+3m+1}{m^2} } +\frac{9m^2(1+3m)}{3m-1} = \\ \\ \frac{( \frac{1}{m^2}-27m)( \frac{1}{m^2}+27m) }{ \frac{(3m-1)^2(9m^2+3m+1)}{m^4} } +\frac{9m^2(1+3m)}{3m-1} = \\ \\ \frac{ \frac{(1-27m^3)(1+27m^3)}{m^4} }{ \frac{(3m-1)^2(9m^2+3m+1)}{m^4} } +\frac{9m^2(1+3m)}{3m-1} =$
$\frac{(1-3m)(1+3m+9m^2)(1+3m)(1-3m+9m^2)}{(3m-1)^2(9m^2+3m+1)} +\frac{9m^2(1+3m)}{3m-1} = \\ \\ \frac{-(1+3m)(1-3m+9m^2)}{3m-1} +\frac{9m^2(1+3m)}{3m-1} = \\ \\ \frac{(1+3m)(9m^2-1+3m-9m^2)}{3m-1} = \frac{(1+3m)(3m-1)}{3m-1} =3m+1$