((n+2)!)/((n-3)!)=30n

Lösung

\frac{( n + 2 ) !}{( n - 3 ) !} = 30 n

n = 0, n = \frac{5 + 129}{2}, n = - \frac{5 + 129}{2}, n = \frac{5 - 129}{2}, n = - \frac{5 - 129}{2}

Schritte zur Lösung

\frac{( n + 2 ) !}{( n - 3 ) !} = 30 n

Vereinfache

\frac{( n + 2 ) !}{( n - 3 ) !} : n (n + 2) (n + 1) (n - 1) (n - 2)

n (n + 2) (n + 1) (n - 1) (n - 2) = 30 n

Löse

n (n + 2) (n + 1) (n - 1) (n - 2) = 30 n : n = 0, n = \frac{5 + 129}{2}, n = - \frac{5 + 129}{2}, n = \frac{5 - 129}{2}, n = - \frac{5 - 129}{2}

n = 0, n = \frac{5 + 129}{2}, n = - \frac{5 + 129}{2}, n = \frac{5 - 129}{2}, n = - \frac{5 - 129}{2}

\frac{3 x - 2}{x - 1} = 2

\frac{4}{8} = \frac{3}{z}

\frac{17}{2 m} - \frac{3}{m} = 1

\frac{2}{x - 4} = \frac{5}{x + 9}

so l v e f or r, r = \frac{c}{1 - r}