Resolva a equação: *
An,4= 2. An,2
Ajudeeem
Soluções para a tarefa
Resposta:
begin{gathered}A_x_,_3 = 4A_x_,_2 \\ \\ \frac{x!}{(x - 3)!} = \frac{4x!}{(x - 2)!} \\ \\ \frac{(x - 2)!}{(x - 3)!} = 4 \\ \\ \frac{(x - 2)\cancel{(x - 3)!}}{\cancel{(x - 3)!}} = 4 \\ \\ x - 2 = 4 \\ x = 8 \end{gathered}
\begin{gathered}A_n_,_2 = 9A_n_,_1 \\ \\ \frac{n!}{(n - 2)!} = \frac{9n!}{(n - 1)!} \\ \\ \frac{(n - 1)!}{(n - 2)!} = 9 \\ \\ \frac{(n - 1)(n - 2)!}{(n - 2)!} = 9 \\ \\ n = 10 \end{gathered}
\begin{gathered} \frac{A_n_,_6\: + \: A_n_,_5}{A_n_,_4} = 9 \\ \\ \frac{ \frac{n!}{(n - 6)!} + \frac{n!}{(n - 5)!} }{ \frac{n!}{(n - 4)!} } = 9 \\ \\ \frac{n(n - 1)(n - 2)(n - 3)(n - 4)(n - 5)\: + \: n(n - 1)(n - 2)(n - 3)(n - 4)}{n(n - 1)(n - 2)(n - 3)} = 9 \\ \\ \frac{n(n - 1)(n - 2)(n - 3)[(n - 4)(n - 5) + n - 4]}{n(n - 1)(n - 2)(n - 3)} = 9 \\ \\ n^2 - 9n + 20 + n - 4 = 9 \\ n^2 - 8n + 7 = 0 \end{gathered}
n' = 7
n" = 1
S = {7}
espero ter ajudado^ ^