Ответ:

[tex]\frac{a+3}{a+1}-\frac{a+1}{a-1}+\frac{6}{a^2-1}=\frac{2}{a^2-1}\\\\[/tex]

[tex]\frac{1}{6}[/tex]

[tex]\frac{18}{c-3}[/tex]

Объяснение:

[tex]\frac{(a+3)(a-1)}{(a-1)(a+1)}-\frac{(a+1)^2}{(a+1)(a-1)}+\frac{6}{(a+1)(a-1)}=\\\\\frac{(a+3)(a-1)-(a+1)^2+6}{(a+1)(a-1)}=\\\\\frac{a^2-a+3a-3-(a^2+2a+1)+6}{(a+1)(a-1)}=\\\\\frac{a^2-a+3a-3-a^2-2a-1+6}{(a+1)(a-1)}=\frac{2}{a^2-1}[/tex]

[tex]\frac{7}{2a-4}-\frac{12}{a^2-4}-\frac{3}{a+2}=\frac{7}{2(a-2)}-\frac{12}{(a-2)(a+2)}-\frac{3}{a+2}=\\\\\frac{7(a+2)}{2(a-2(a+2)}-\frac{12\cdot 2}{2(a-2)(a+2)}-\frac{3\cdot 2(a-2)}{2(a-2)(a+2)}=\\\\\frac{7(a+2)-12\cdot 2-3\cdot 2(a-2)}{2(a-2)(a+2)}=\frac{7a+14-24-6a+12}{2(a-2)(a+2)}=\\\\\frac{a+2}{2(a-2)(a+2)}=\frac{1}{2(a-2)}\\\\\frac{1}{2\cdot(5-2)}=\frac{1}{2\cdot3}=\frac{1}{6}[/tex]

[tex]\frac{c^2+9}{c-3}-c-3=\frac{c^2+9}{c-3}-\frac{(c+3)(c-3)}{c-3}=\\\\ \frac{c^2+9-(c+3)(c-3)}{c-3}=\frac{c^2+9-(c^2-9)}{c-3}=\frac{c^2+9-c^2+9}{c-3}=\frac{18}{c-3}[/tex]