Question

РЕШИТЕ УРАВНЕНИЯ
$c \frac{2}{n} = 28$
$c \frac{n - 3}{n} = 20$
$c \frac{m}{30} = 435$
С подробным объяснением пжжж!!!​

Answer 1

$\displaystyle\bf\\1)\\\\C_{n} ^{2} =28\\\\\\\frac{n!}{2!\cdot(n-2)!} =28\\\\\\\frac{(n-2)!\cdot(n-1)\cdot n}{1\cdot 2\cdot(n-2)!} =28\\\\\\n(n-1)=56\\\\n^{2} -n-56=0\\\\Teorema \ \ Vieta:\\\\n_{1} =8\\\\n_{2} =-7<0-neyd\\\\Otvet:8\\\\\\2)\\\\C_{n} ^{n-3} =20\\\\\\\frac{n!}{(n-3)!\cdot(n-n+3)!} =20\\\\\\\frac{(n-3)!\cdot(n-2)\cdot(n-1)\cdot n}{(n-3)!\cdot3!} =20\\\\\\\frac{(n-2)(n-1)n}{1\cdot 2\cdot3} =20$

$\displaystyle\bf\\(n-2)(n-1)n=120\\\\n_{1} =4 \ \ ; \ \ n_{2} =5 \ \ ; \ n_{3} =6\\\\Otvet:4 \ ; \ 5 \ ; \ 6$

Третье задание верно записано ?