giải phương trình x! - (x-1)! / (n+1)! =1/6 n!/(n-2)! -n!(n-1)! =3 n^3 +n!/(n-2)!=10

Đáp án:

Giải thích các bước giải: \[\begin{array}{l} \frac{{n! - (n - 1)!}}{{(n + 1)!}} = \frac{1}{6}(dk:n \ge 1;n \in N)\\ \Leftrightarrow \frac{{n!}}{{(n + 1)!}} - \frac{{(n - 1)!}}{{(n + 1)!}} = \frac{1}{6}\\ \Leftrightarrow \frac{1}{{n + 1}} - \frac{1}{{n(n + 1)}} = \frac{1}{6}\\ \Leftrightarrow n = 5 \to tm\\ \frac{{n!}}{{(n - 2)!}} - \frac{{n!}}{{(n - 1)!}} = 3(dk:n \ge 2;n \in N)\\ \Leftrightarrow n(n - 1) - n = 3\\ \Leftrightarrow \left[ \begin{array}{l} n = 3 \to tm\\ n = - 1 \to loai \end{array} \right.\\ {n^3} + \frac{{n!}}{{(n - 2)!}} = 10(dk:n \ge 2;n \in N)\\ \Leftrightarrow {n^3} + n(n - 1) = 10\\ \Leftrightarrow n = 2 \to tm \end{array}\]

$\begin{array}{l}
 \dfrac{{n! - (n - 1)!}}{{(n + 1)!}} = \dfrac{1}{6}(dk:n \ge 1;n \in N)\\
  \Leftrightarrow \dfrac{{n!}}{{(n + 1)!}} - \dfrac{{(n - 1)!}}{{(n + 1)!}} = \dfrac{1}{6}\\
  \Leftrightarrow \dfrac{1}{{n + 1}} - \dfrac{1}{{n(n + 1)}} = \dfrac{1}{6}\\
  \Leftrightarrow n = 5 \to tm\\
 \dfrac{{n!}}{{(n - 2)!}} - \dfrac{{n!}}{{(n - 1)!}} = 3(dk:n \ge 2;n \in N)\\
  \Leftrightarrow n(n - 1) - n = 3\\
  \Leftrightarrow \left[ \begin{array}{l}
 n = 3 \to tm\\
 n =  - 1 \to loai
 \end{array} \right.\\
 {n^3} + \dfrac{{n!}}{{(n - 2)!}} = 10(dk:n \ge 2;n \in N)\\
  \Leftrightarrow {n^3} + n(n - 1) = 10\\
  \Leftrightarrow n = 2 \to tm
 \end{array}$