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Решите задачу:

$\frac{4}{y+2} - \frac{3}{y-2} + \frac{12}{4- y^{2} } = \frac{4(y-2)-3(y+2)+12}{(y+2)(y-2)} = \frac{4y-8-3y-6+12}{(y+2)(y-2)} = \\ \frac{y-2}{(y+2)(y-2)} = \frac{1}{y+2} \\ \\ \\ \\. \\ \frac{a}{a-6} - \frac{3}{a+6} + \frac{ a^{2} }{36-a^{2}} = \frac{a}{a-6}- \frac{3}{a+6} - \frac{a^{2} }{(a-6)(a+6)}= \frac{a(a+6)-3(a-6)-a^{2} }{(a-6)(a+6)} = \\ \\ \frac{a^{2} +6a-3a+18-a^{2} }{(a-6)(a+6)} = \frac{3a+18}{(a-6)(a+6)} = \frac{3(a+6)}{(a-6)(a+6)}= \frac{3}{a-6}$