Question

alguém pode mostrar a resolução pra mim?
Resposta: E

Answer 1

$\frac{x+3}{x^{2}-1} = \frac{A}{x-1}+\frac{B}{x+1} \\\\ MMC = (x-1)(x+1) \\\\ \frac{x+3}{(x+1)(x-1)} = \frac{A(x+1)+B(x-1)}{(x+1)(x+1)} \\\\ x+3 = A(x+1)+B(x-1) \\\\ x+3 = Ax+A+Bx-B \\\\ x+3 = Ax+Bx+A-B \\\\ x+3 = (A+B)x+(A-B) \\\\\ \left\{\begin{matrix} A+B=1 & \\ A-B = 3 & \end{matrix}\right.$


Resolvendo o sistema:

$\left\{\begin{matrix} A+B=1 & \\ A-B = 3 & \end{matrix}\right. \\\\ 2A = 4 \Rightarrow \boxed{A = 2} \\\\ A+B = 1 \ \Rightarrow \ 2+B = 1 \ \Rightarrow \ \boxed{B = -1}$

Alternativa E.