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Решите задачу:

$\frac{9a-4}{a+7} - \frac{44-16a}{a^2+5a-14} = \frac{9a-4}{a+7} - \frac{44-16a}{a^2+7a-2a-14}=$

$= \frac{9a-4}{a+7} - \frac{44-16a}{a*a+a*7+(-2)*a+(-2)*7}= \frac{9a-4}{a+7} - \frac{44-16a}{a*(a+7)+(-2)*(a+7)}=$

$= \frac{9a-4}{a+7}* \frac{1}{1}-\frac{44-16a}{(a-2)*(a+7)}=$

$= \frac{9a-4}{a+7}* \frac{a-2}{a-2}-\frac{44-16a}{(a-2)*(a+7)}=$

$= \frac{(9a-4)(a-2)-(44-16a)}{(a+7)(a-2)}=$

$= \frac{9a*(a-2)-4*(a-2)-44+16a}{(a+7)(a-2)} = \frac{9a*a-9a*2-4*a+(-4)*(-2)-44+16a}{(a+7)(a-2)}=$

$= \frac{9a^2-18a-4a+8-44+16a}{(a+7)(a-2)} = \frac{9a^2-18a-4a+16a+8-44}{(a+7)(a-2)}=$

$= \frac{9a^2-6a-36}{(a+7)(a-2)} = \frac{3*3a^2-3*2a-3*12}{(a+7)(a-2)}= \frac{3(3a^2-2a-12)}{(a+7)(a-2)}$