Помогите пожалуйста решить 78 номер

Решите задачу:

$\frac{10-x}{x-7}+\frac{2x-11}{7-x}=\frac{10-x}{x-7}-\frac{2x-11}{x-7}=\\ \\=\frac{10-x-(2x-11)}{x-7}=\frac{10-x-2x+11}{x-7}=\frac{21-3x}{x-7}=-\frac{3*(7-x}{7-x}=-3\\$
$\\\frac{11y-8}{y^2-25}+\frac{12+7y}{25-y^2}=\frac{11y-8}{y^2-25}-\frac{12+7y}{y^2-25}=\\=frac{11y-8-(12+7y)}{(y-5)(y+5)}=\frac{11y-8-12-7y}{(y-5)(y+5)}=\\=\frac{4(y-5)}{(y-5)(y+5)}=\frac{4}{y+5}\\$
$\\\frac{a^2-13a}{a^2-36} - \frac{36+a}{36-a^2} = \frac{a^2-13a}{a^2-36} + \frac{36+a}{a^2-36}=\\\frac{a^2-13a+36+a}{(a-6)(a+6)}=\frac{a^2-12a+36}{(a-6)(a+6)}=\\= \\\frac{(a-6)^2}{(a-6)(a+6)}=\frac{a-6}{a+6}\\$

$\\\frac{b^2+8b}{49-b^2}-\frac{6b+49}{b^2-49}=\frac{b^2+8b}{49-b^2}+\frac{6b+49}{49-b^2}=\\=\frac{b^2+8b+6b+49}{(7-b)(7+b)}=\frac{b^2+14b+49}{(7-b)(7+b)}=\\= \frac{(b+7)^2}{(7-b)(b+7)}=\frac{b+7}{7-b}\\$