Question

7/x-1 - 8/x+1 = 9/x²-1 é um divisor de​

Answer 1

$\displaystyle \sf \frac{7}{x-1}-\frac{8}{x+1}=\frac{9}{x^2-1} \\\\\\\ \frac{7}{(x-1)/(x+1)}-\frac{8}{(x+1)/(x-1)}=\frac{9}{x^2-1} \\\\\\ \frac{7(x+1)-8(x-1)}{(x-1)(x+1)}=\frac{9}{x^2-1} \\\\\\ \frac{7x+7-8x+8}{x^2-1}=\frac{9}{x^2-1} \\\\\\ -x+15=9 \\\\ -x = 9-15 \\\\ \huge\boxed{\sf \ x = 6\ }\checkmark$