Решите, то что-то не так выходит 1) (с+3/с-3 - с/с+3)•с-3/с+1 2) (14+а²/а²-4 - а-4/а+2)•а-2/6

[latex]( \frac{c+3}{c-3} - \frac{c}{c+3} ) \frac{c-3}{c+1} =( \frac{(c+3)^2}{(c-3)(c+3)} - \frac{c(c-3)}{(c+3)(c-3)} ) \frac{c-3}{c+1} = \frac{(c+3)^2-c (c-3)}{c^2-9} \frac{c-3}{c+1}= \\ \frac{c^2+6c+9-c^2+3c}{c^2-9} \frac{c-3}{c+1}= \frac{9c+9}{c^2-9} \frac{c-3}{c+1}=\frac{9 (c+1)(c-3)}{(c-3)(c+3)(c+1)}= \frac{9}{c+3} [/latex] [latex]( \frac{14+a^2}{a^2-4} - \frac{a-4}{a+2} ) \frac{a-2}{6} =( \frac{14+a^2}{a^2-4} - \frac{(a-4)(a-2)}{(a+2)(a-2)} ) \frac{a-2}{6}=\frac{14+a^2-(a-4)(a-2)}{a^2-4} \frac{a-2}{6}= \\ =\frac{14+a^2-(a^2-2a-4a+8)}{a^2-4} \frac{a-2}{6}=\frac{14+a^2-a^2+6a-8}{a^2-4} \frac{a-2}{6}=\frac{6+6a}{(a-2)(a+2)} \frac{a-2}{6}= \\ = \frac{a+1}{a+2} [/latex]