Question

ПОМОГИТЕ ПОЖАЛУЙСТА 45 БАЛЛОВ

Answer 1

$\frac{3a^{2}+6 }{a^{3}+1 } -\frac{3}{a^{2}-a+1 }-\frac{1}{a+1}= \frac{3a^{2}+6 }{(a+1)(a^{2}-a+1)} -\frac{3}{a^{2}-a+1 }-\frac{1}{a+1}=\\\\=\frac{3a^{2}+6-3\cdot(a+1)-1\cdot(a^{2}-a+1)}{(a+1)(a^{2}-a+1)}=\frac{3a^{2}+6-3a-3-a^{2}+a-1}{(a+1)(a^{2}-a+1) }=\frac{2a^{2}-2a+2 }{(a+1)(a^{2}-a+1) }=\\\\=\frac{2(a^{2}-a+1) }{(a+1)(a^{2}-a+1) }=\frac{2}{a+1} \\\\a=-1,4\\\\\frac{2}{a+1}=\frac{2}{-1,4+1} =\frac{2}{-0,4}=\boxed{-2,5}$