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Решите задачу:

$( \frac{6a+1}{a^2-6a}+ \frac{6a-1}{a^2+6a} )\cdot \frac{a^2-36}{a^2+1} = ( \frac{6a+1}{a(a-6)}+ \frac{6a-1}{a(a+6)} )\cdot \frac{a^2-36}{a^2+1} = \\\ = \frac{(6a+1)(a+6)+(6a-1)(a-6)}{a(a-6)(a+6)}\cdot \frac{a^2-36}{a^2+1} = \frac{6a^2+36a+a+6+6a^2-36a-a+6}{a(a-6)(a+6)}\cdot \frac{a^2-36}{a^2+1} = \\\ = \frac{12a^2+12}{a(a-6)(a+6)}\cdot \frac{(a-6)(a+6)}{a^2+1} = \frac{12(a^2+1)(a-6)(a+6)}{a(a-6)(a+6)(a^2+1)}= \frac{12}{a}$

$\frac{x^2-10x+25}{35-7x} = \frac{(5-x)^2}{7(5-x)} = \frac{5-x}{7}$