Хелп ми с алгеброй 527 срочно

Решите задачу:

$\displaystyle 1)..\frac{a^{2}+1}{a-3}- \frac{10}{a-3}= \frac{a^{2}-9}{a-3}= \frac{(a-3)(a+3)}{a-3}=a+3; \\ \\npu:a=97 \\\\ a+3=97+3=100 \\ \\ \\ \\ 2).. \frac{x+7}{x^{2}-25}- \frac{2x+2}{x^{2}-25}= \frac{x+7-2x-2}{(x-5)(x+5)}= -\frac{x-5}{(x-5)(x+5)}= \\ \\ =-\frac{1}{x+5}; \\ \\npu:x=-5,1 \\ \\- \frac{1}{x+5}=- \frac{1}{5-5,1}=1:0,1=10$