Докажите тождество:
1) $\frac{1}{a^2 + 12a + 36} + \frac{2}{36 - a^2} + \frac{1}{a^2 - 12a + 36} = \frac{144}{(a^2 - 36)^2}$;
2) $\frac{a^2}{(a - b)(a - c)} + \frac{b^2}{(b - a)(b - c)} + \frac{c^2}{(c - a)(c - b)} = 1$.
$\frac{1}{a^2 + 12a + 36} + \frac{2}{36 - a^2} + \frac{1}{a^2 - 12a + 36} = \frac{1}{(a + 6)^2} + \frac{2}{(6 - a)(6 + a)} + \frac{1}{(a - 6)^2} = \frac{1}{(a + 6)^2} - \frac{2}{(a - 6)(a + 6)} + \frac{1}{(a - 6)^2} = \frac{(a - 6)^2 - 2(a - 6)(a + 6) + (a + 6)^2}{(a - 6)^2(a + 6)^2} = \frac{a^2 - 12a + 36 - 2(a^2 - 36) + a^2 + 12a + 36}{(a - 6)^2(a + 6)^2} = \frac{2a^2 + 36 - 2a^2 + 72 + 36}{((a - 6)(a + 6))^2} = \frac{144}{(a^2 - 36)^2}$
$\frac{a^2}{(a - b)(a - c)} + \frac{b^2}{(b - a)(b - c)} + \frac{c^2}{(c - a)(c - b)} = \frac{a^2}{(a - b)(a - c)} - \frac{b^2}{(a - b)(b - c)} + \frac{c^2}{(a - c)(b - c)} = \frac{a^2(b - c) - b^2(a - c) + c^2(a - b)}{(a^2 - ab - ac + bc)(b - c)} = \frac{a^2b - a^2c - ab^2 + b^2c + ac^2 - bc^2}{a^2b - ab^2 - abc + b^2c - a^2c + abc + ac^2 - bc^2} = \frac{a^2b - a^2c - ab^2 + b^2c + ac^2 - bc^2}{a^2b - a^2c - ab^2 + b^2c + ac^2 - bc^2} = 1$
Пожауйста, оцените решение