Помогите даю 10 баллов с 25 по 26

Решите задачу:

$25)\; \; \; \left\{\begin{array}{c}2q-4p=-9\\2t-4q=-7\\2p-4t=2\end{array}\right \; \; \; \oplus \\----------------------\\(-2q-4p)+(2t-4q)+(2p-4t)=-9-7+2\\\\-2q-2p-2t=-14\\\\-2(q+p+t)=-14\\\\p+q+t=7$

$26)\; \; \frac{1}{x^2-5x-6} =\frac{a}{x-6}+\frac{b}{x+1}\\\\x^2-5x-6=0\\\\x_1=6\; ,\; \; x_2=-1\; \; (teorema\; Vieta)\; \; \to \\\\x^2-5x-6=(x-6)(x+1)\\\\ \frac{1}{(x-6)(x+1)}= \frac{a}{x-6}+\frac{b}{x+1}= \frac{a(x+1)+b(x-6)}{(x-6)(x+1)} \\\\1=a(x+1)+b(x-6)\\\\1=ax+a+bx-6b\\\\1=(a+b)\cdot x+(a-6b)\\\\0\cdot x+1\cdot x^0=(a+b)\cdot x+(a-6b)\cdot x^0\\\\x\; |\; a+b=0\quad \to \; \; a=-b\\\\x^0|\; a-6b=1\quad \to \; \; -7b=1\; ,\; b=-\frac{1}{7}\\\\a=\frac{1}{7}\\\\Otvet:\; \; a= \frac{1}{7}\; ,\; b=-\frac{1}{7}\; .$