Question

(5a ^ 2 - 2a + 8)/(a ^ 3 + 1) - 4/(a + 1) - 3/(a ^ 2 - a + 1) = 1/(a + 1) ПОМОГИТЕ ПЖ ДАЮ 100 БАЛОВ

Answer 1

Ответ:

$\frac{5a ^ 2 - 2a + 8}{a ^ 3 + 1} - \frac{4}{a + 1} - \frac{3}{a ^ 2 - a + 1} =\frac{ 1}{a + 1}$

Объяснение:

$\frac{5a ^ 2 - 2a + 8}{a ^ 3 + 1} - \frac{4}{a + 1} - \frac{3}{a ^ 2 - a + 1} =\frac{5a ^ 2 - 2a + 8}{(a + 1)(a^2 - a + 1)} - \frac{4(a^2 - a + 1)}{(a + 1)(a^2 - a + 1)} - \frac{3(a+1)}{(a + 1)(a^2 - a + 1)} =\\\\\frac{5a ^ 2 - 2a + 8 -4(a^2 - a + 1) - 3(a+1)}{(a + 1)(a^2 - a + 1)} =\frac{5a ^ 2 - 2a + 8 -4a^2 +4a - 4 - 3a-3}{(a + 1)(a^2 - a + 1)} =\\\\\frac{a^2 - a + 1}{(a + 1)(a^2 - a + 1)} =\frac{ 1}{a + 1}$