Докажите тождество:
1) $\frac{a + 3}{a^2 - 3a} + \frac{a - 3}{3a + 9} + \frac{12}{9 - a^2} = \frac{a - 3}{3a}$;
2) $\frac{b - 4}{2a - 1} - \frac{b^2 - 2b - 24}{2ab - 4 - b + 8a} = \frac{2}{2a - 1}$.
$\frac{a + 3}{a^2 - 3a} + \frac{a - 3}{3a + 9} + \frac{12}{9 - a^2} = \frac{a - 3}{3a}$
$\frac{a + 3}{a^2 - 3a} + \frac{a - 3}{3a + 9} - \frac{12}{a^2 - 9} = \frac{a - 3}{3a}$
$\frac{a + 3}{a(a - 3)} + \frac{a - 3}{3(a + 3)} - \frac{12}{(a - 3)(a + 3)} = \frac{a - 3}{3a}$
$\frac{3(a + 3)^2 + a(a - 3)^2 - 3a * 12}{3a(a - 3)(a + 3)} = \frac{a - 3}{3a}$
$\frac{3(a^2 + 6a + 9) + a(a^2 - 6a + 9) - 36a}{3a(a - 3)(a + 3)} = \frac{a - 3}{3a}$
$\frac{3a^2 + 18a + 27 + a^3 - 6a^2 + 9a - 36a}{3a(a - 3)(a + 3)} = \frac{a - 3}{3a}$
$\frac{a^3 - 3a^2 - 9a + 27}{3a(a - 3)(a + 3)} = \frac{a - 3}{3a}$
$\frac{(a^3 - 3a^2) - (9a - 27)}{3a(a - 3)(a + 3)} = \frac{a - 3}{3a}$
$\frac{a^2(a - 3) - 9(a - 3)}{3a(a - 3)(a + 3)} = \frac{a - 3}{3a}$
$\frac{(a - 3)(a^2 - 9)}{3a(a^2 - 9)} = \frac{a - 3}{3a}$
$\frac{a - 3}{3a} = \frac{a - 3}{3a}$
$\frac{b - 4}{2a - 1} - \frac{b^2 - 2b - 24}{2ab - 4 - b + 8a} = \frac{2}{2a - 1}$
$\frac{b - 4}{2a - 1} - \frac{b^2 - 2b - 24}{(2ab - b) + (8a - 4)} = \frac{2}{2a - 1}$
$\frac{b - 4}{2a - 1} - \frac{b^2 - 2b - 24}{b(2a - 1) + 4(2a - 1)} = \frac{2}{2a - 1}$
$\frac{b - 4}{2a - 1} - \frac{b^2 - 2b - 24}{(2a - 1)(b + 4)} = \frac{2}{2a - 1}$
$\frac{(b - 4)(b + 4) - (b^2 - 2b - 24)}{(2a - 1)(b + 4)} = \frac{2}{2a - 1}$
$\frac{b^2 - 4b + 4b - 16 - b^2 + 2b + 24}{(2a - 1)(b + 4)} = \frac{2}{2a - 1}$
$\frac{2b + 8}{(2a - 1)(b + 4)} = \frac{2}{2a - 1}$
$\frac{2(b + 4)}{(2a - 1)(b + 4)} = \frac{2}{2a - 1}$
$\frac{2}{2a - 1} = \frac{2}{2a - 1}$
Пожауйста, оцените решение
