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Решите задачу:

$\frac{a^2}{a+3}-\frac{9}{a+3}=\frac{a^2-9}{(a+3}=\frac{(a-3)(a+3)}{a+3}=a-3\\\\\frac{t}{t^2-16}-\frac{4}{t^2-16}=\frac{t-4}{t^2-16}=\frac{t-4}{(t-4)(t+4)}\\\\\frac{m^2}{(m-5)^2}-\frac{25}{(m-5)^2}=\frac{m^2-25}{(m-5)^2}=\frac{(m-5)(m+5)}{(m-5)^2}=\frac{m+5}{m-5}$

$\frac{5x+9}{x^2-1}-\frac{4x+8}{x^2-1}=\frac{5x+9-4x-8}{x^2-1}=\frac{x+1}{(x-1)(x+1)}=\frac{1}{x-1}\\\\\frac{b^2}{b+10}+\frac{20b+100}{b+10}=\frac{b^2+20b+100}{b+10}=\frac{(b+10)^2}{b+10}=b+10\\\\\frac{c^2}{c-7}-\frac{14c-49}{c-7}=\frac{c^2-14c+49}{c-7}=\frac{(c-7)^2}{c-7}=c-7$