Докажите тождество:
$\frac{3a^2 + 24}{a^3 + 8} - \frac{6}{a^2 - 2a + 4} - \frac{1}{a + 2} = \frac{2}{a + 2}$.
$\frac{3a^2 + 24}{a^3 + 8} - \frac{6}{a^2 - 2a + 4} - \frac{1}{a + 2} = \frac{2}{a + 2}$
$\frac{3a^2 + 24}{(a + 2)(a^2 - 2a + 4)} - \frac{6}{a^2 - 2a + 4} - \frac{1}{a + 2} = \frac{2}{a + 2}$
$\frac{3a^2 + 24 - 6(a + 2) - (a^2 - 2a + 4)}{(a + 2)(a^2 - 2a + 4)} = \frac{2}{a + 2}$
$\frac{3a^2 + 24 - 6a - 12 - a^2 + 2a - 4}{(a + 2)(a^2 - 2a + 4)} = \frac{2}{a + 2}$
$\frac{2a^2 - 4a + 8}{(a + 2)(a^2 - 2a + 4)} = \frac{2}{a + 2}$
$\frac{2(a^2 - 2a + 4)}{(a + 2)(a^2 - 2a + 4)} = \frac{2}{a + 2}$
$\frac{2}{a + 2} = \frac{2}{a + 2}$
Пожауйста, оцените решение
