Question

помагите срочно!!!!!!!!!​

Answer 1

1.

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

2. Приравняем $a^{2}+2a+1 =0$

$D = b^{2} - 4ac$

$D = 4 - 4$

$D = 0$

$a = \frac{-2}{2}$

$a = -1$

$a^{2}+2a+1 = (a+1)^{2}$

3. Подставим

$\frac{(a+1)^{2}}{a(a-1)(a+1)} = \frac{(a+1)}{a(a-1)}$