Задание 18 стр 106

a) $\frac{4}{x+1}-\frac{3}{x-1}+\frac{8}{x^2-1}=\frac{4(x-1)-3(x+1)+8}{x^2-1}=\frac{4x-4-3x-3+8}{x^2-1}-\frac{x+1}{x^2-1}=\frac{x+1}{(x-1)(x+1)}=\frac{1}{x-1}\text{. }$

$\begin{array}{ccc}\hfill b)& \frac{a}{a-2}-\frac{1}{a+2}+\frac{a^2}{4-a^2}=\frac{a}{a-2}-\frac{1}{a+2}-\frac{a^2}{a^2-4}=\hfill & \hfill \\ \hfill & =\frac{a(a+2)-(a-2)}{(a-2)(a+2)}-\frac{a^2}{a^2-4}=\frac{a^2+2a-a+2-a^2}{a^2-4}=\hfill \\ \hfill & =\frac{a+2}{(a-2)(a+2)}=\frac{1}{a-2}.\hfill \end{array}$
c) $\frac{4k-12}{(2+k-12}+\frac{k}{k+4}=\frac{4(k-3)}{(k+4)(k-3)}+\frac{k}{k+4}=\frac{4}{k+4}+\frac{k}{k+4}=\frac{4+k}{k+4}=1$.
d) $\frac{2}{x^2-3x}-\frac{1}{x^2+3x}-\frac{x+1}{x^2-9}=\frac{2}{x(x-3)}-\frac{1}{x(x+3)}-\frac{x+1}{(x-3)(x+3)}=\frac{2(x+3)-(x-3)-x(x+1)}{x(x-3)(x+3)}=\frac{2x+6-x+3-x^2-x}{x(x-3)(x+3)}=\frac{-x^2+9}{x(x-3)(x+3)}=$
$=\frac{-(x-3)(x+3)}{x(x-3)(x+3)}=-\frac{1}{x}.$
e) $\frac{2}{x^2-x}+\frac{1}{x^2+x}-\frac{x+3}{x^2-1}=\frac{2}{x(x-1)}+\frac{1}{x(x+1)}-\frac{x+3}{(x-1)(x+1)}=\frac{2(x+1)+x-1-x(x+3)}{x(x-1)(x+1)}=\frac{2x+2+x-1-x^2-3x}{x(x-1)(x+1)}=$
$\frac{-x^2+1}{x(x-1)(x+1)}=-\frac{x^2-1}{x(x-1)(x+1)}=-\frac{1}{x}$