Задание 14 стр 106

$\begin{array}{ccc}\hfill a)& \frac{x+5}{5x-25}-\frac{3x+5}{x^2-25}=\frac{x+5}{5(x-5)}-\frac{3x+5}{(x-5)(x+5)}=\frac{{(x+5)}^2-5(3x+5)}{5(x-5)(x+5)}=\frac{x^2+10x+25-15x-25}{5\left(x{x}^2-25\right)}=\frac{x^2-5x}{5(x-5)(x+5)}=\frac{x(x-5)}{5(x-5)(x+5)}=\hfill & \hfill \\ \hfill =& \frac{x}{5(x+5)}.\hfill \end{array}$
$\begin{array}{ccc}\hfill b)& \frac{y-8}{4y-16}+\frac{4}{y^2-4y}=\frac{y-8}{4(y-4)}+\frac{4}{y(y-4)}=\frac{(y-8)y+16}{4y(y-4)}=\hfill & \hfill \\ \hfill =& \frac{y^2-8y+16}{4y(y-4)}=\frac{{(y-4)}^2}{4y(y-4)}=\frac{y-4}{4y}.\hfill \end{array}$ $\begin{array}{ccc}\hfill c)& \frac{b-6}{4-b^2}+\frac{2}{2b-b^2}=\frac{b-6}{(2-b)(2+b)}+\frac{2}{b(2-b)}=\frac{b^2-6b+4+2b}{b(2-b)(2+b)}=\hfill & \hfill \\ \hfill & =\frac{b^2-4b+4}{b(2-b)(2+b)}=\frac{{(b-2)}^2}{b(2-b)(2+b)}=\frac{{(2-b)}^2}{b(2-b)(2+b)}=\frac{2-b}{b(2+b)}=\hfill \\ \hfill =& \frac{2-b}{2b+b^2}.\hfill \end{array}$