Question

помогите пожалуйста решить пример​

Answer 1

а)

$\frac{n+3}{2n+2} -\frac{n+1}{2n-2} +\frac{3}{n^{2}-1} =\frac{n+3}{2(n+1)} -\frac{n+1}{2(n-1)} +\frac{3}{(n-1)(n+1)} =\frac{(n+3)(n-1)}{2(n-1)(n+1)} -\frac{(n+1)(n+1)}{2(n-1)(n+1)} +\\\\+\frac{6}{2(n-1)(n+1)} =\frac{n^{2}+2n-3-n^{2}-2n-1+6}{2(n-1)(n+1)} =\frac{2}{2(n-1)(n+1)} =\frac{1}{n^{2}-1}$

б)

$\frac{2a^{2}+7a+9}{a^{3}-1} +\frac{4a+3}{a^{2}+a+1} =\frac{2a^{2}+7a+9}{(a-1)(a^{2}+a+1)} +\frac{(4a+3)(a-1)}{(a-1)(a^{2}+a+1)} =\frac{2a^{2}+7a+9+4a^{2}-a-3}{(a-1)(a^{2}+a+1)} =\\\\=\frac{6a^{2}+6a+6}{(a-1)(a^{2}+a+1)} =\frac{6(a^{2}+a+1)}{(a-1)(a^{2}+a+1)} =\frac{6}{a-1}$