Question

resolva as equações (n+1)!+2(n-1)!/(n+1)-2(n-1)!=7/5​

Answer 1

Resposta:

Leia abaixo

Explicação passo-a-passo:

$\boxed{\frac{(n+1)!+2(n-1)!}{(n+1)!-2(n-1)!} = \frac{7}{5} }$

$\boxed{\frac{(n+1)!+2(n-1).n.(n+1)!}{(n+1)!-2(n-1).n.(n+1)!} = \frac{7}{5} }$

$\boxed{\frac{(n+1)!( 1 +2(n-1).n)}{(n+1)!(1 -2(n-1).n)} = \frac{7}{5} }$

$\boxed{\frac{1 +2n^2-2n}{1 -2n^2- 2n} = \frac{7}{5} }$

$5 + 10n^2 - 10n = 7 - 14n^2 - 14n$

$24n^2 + 4n - 2 = 0$

$\boxed{\boxed{x' = \frac{-1-\sqrt{13} }{12}}}$

$\boxed{\boxed{x'' = \frac{-1+\sqrt{13} }{12}}}$