Question

298. Сократите дробь:​

Answer 1

$a)\ \ \dfrac{A_5^2}{3!}=\dfrac{5\cdot 4}{1\cdot 2\cdot 3}=\dfrac{10}{3}\\\\\\b)\ \ \dfrac{A_6^3}{4!}=\dfrac{6\cdot 5\cdot 4}{1\cdot 2\cdot 3\cdot 4}=5\\\\\\c)\ \ \dfrac{A_{n+1}^3}{A_{n}^2}=\dfrac{(n+1)\cdot n\cdot (n-1)}{n\cdot (n-1)}=n+1\\\\\\d)\ \ \dfrac{A_{n+3}^3}{A_{n+1}^2}=\dfrac{(n+3)(n+2)(n+1)}{(n+1)\cdot n}=\dfrac{(n+3)(n+2)}{n}=\dfrac{n^2+5n+6}{n}$