Question

Calcule p,p>3,sendo dado:

Answer 1

$\frac{ \frac{(p-1)!}{2!(p-3)!} + \frac{(p-1)!}{3!(p-4)!} }{ \frac{p!}{2!(p-2)!} - \frac{(p-1)!}{1!(p-2)!} } = \frac{5}{3}$

$\frac{ \frac{(p-1).(p-2).(p-3)!}{2.(p-3)!} + \frac{(p-1).(p-2).(p-3).(p-4)!}{3.2.(p-4)!} }{ \frac{p.(p-1).(p-2)!}{2.(p-2)!} - \frac{(p-1).(p-2)!}{(p-2)!} } = \frac{5}{3}$

$\frac{ \frac{(p-1).(p-2)}{2} + \frac{(p-1).(p-2).(p-3)}{6} }{ \frac{p.(p-1)}{2} - \frac{(p-1)}{1} } = \frac{5}{3}$

$\frac{ \frac{3.(p-1).(p-2)}{6} + \frac{(p-1).(p-2).(p-3)}{6} }{ \frac{p.(p-1)}{2} - \frac{2.(p-1)}{2} } = \frac{5}{3}$

$\frac{ \frac{3.(p-1).(p-2)+(p-1).(p-2).(p-3)}{6} }{ \frac{p.(p-1)-2.(p-1)}{2} } = \frac{5}{3}$

$\frac{ \frac{(p-1).(p-2).[3+(p-3)]}{6} }{ \frac{(p-1).(p-2)}{2} } = \frac{5}{3}$

$\frac{(p-1).(p-2).p}{6} . \frac{2}{(p-1).(p-2)} = \frac{5}{3}$

$\frac{p}{6} . \frac{2}{1} = \frac{5}{3}$

$\frac{p}{3} . \frac{1}{1} = \frac{5}{3}$

$\frac{p}{3} = \frac{5}{3}$

$\frac{p}{1} = \frac{5}{1}$

$p=5$

Answer 2

Resposta:

p=5

Explicação passo a passo:

vou usar matriz, pois esse símbolo não tem aqui

na parte de cima, usando Stifel

$\left[\begin{array}{ccc}p-1+1\\2+1\end{array}\right] \\\left[\begin{array}{ccc}p\\3\end{array}\right]$

na parte de baixo, usando Stifel também

$\left[\begin{array}{ccc}p-1\\1\end{array}\right] + \left[\begin{array}{ccc}p-1\\2\end{array}\right] = \left[\begin{array}{ccc}p-1+1\\1+1 \end{array}\right] = \left[\begin{array}{ccc}p\\2\end{array}\right]$

logo

$\left[\begin{array}{ccc}p-1\\1\end{array}\right] + \left[\begin{array}{ccc}p-1\\2\end{array}\right] = \left[\begin{array}{ccc}p\\2\end{array}\right] \\\\\left[\begin{array}{ccc}p-1\\2\end{array}\right] = \left[\begin{array}{ccc}p\\2\end{array}\right] - \left[\begin{array}{ccc}p-1\\1\end{array}\right]$

substituindo

$\frac{\left[\begin{array}{ccc}p\\3\\\end{array}\right] }{\left[\begin{array}{ccc}p-1\\2\end{array}\right] } = \frac{5}{3}$

multiplica em cruz

$\left[\begin{array}{ccc}p\\3\end{array}\right] 3 = \left[\begin{array}{ccc}p-1\\2\end{array}\right] 5\\\\\frac{3 (p)(p-1)}{3.2.1} =\frac{5(p-1)}{2.1}$

simplifica

$\frac{ (p)}{1} =\frac{5}{1}\\p=5$