Задача 75527 Из данной пропорции найти...
Условие
Решение
[m]С^{y+1}_{x}=\frac{x!}{(x–(y+1))!\cdot (y+1)!}=\frac{x!}{(x–y-1)!\cdot (y+1)!}[/m]
[m]С^{y}_{x}=\frac{x!}{(x-y)!\cdot y!}[/m]
[m]С^{y+2}_{x}:С^{y+1}_{x}=\frac{x!}{(x-y-2)!\cdot (y+2)!}:\frac{x!}{(x–y-1))!\cdot (y+1)!}=\frac{x!}{(x-y-2)!\cdot (y+2)!}\cdot\frac{(x–y-1)!\cdot (y+1)!}{x!}=\frac{(x–y-2)!\cdot (x-y-1)\cdot (y+1)!}{(x-y-2)!\cdot (y+1)!\cdot (y+2)}=\frac{ x-y-1}{y+2}[/m]
По условию:
[m]С^{y+2}_{x}:С^{y+1}_{x}=3:5[/m]
⇒ [m]\frac{ x-y-1}{y+2}=\frac{3}{5}[/m] ⇒ [m]5(x-y-1)=3(y+2)[/m] ⇒ [m] 5x-8y=11[/m]
[m]С^{y+1}_{x}:С^{y}_{x}=\frac{x!}{(x–y-1)!\cdot (y+1)!}:\frac{x!}{(x-y)!\cdot y!}=\frac{x!}{(x–y-1)!\cdot (y+1)!}\cdot \frac{(x-y)!\cdot y!}{x!}=\frac{x-y}{y+1}[/m]
По условию:
[m]С^{y+1}_{x}:С^{y}_{x}=5:5[/m]
⇒ [m]\frac{ x-y}{y+1}=1[/m] ⇒ [m]x-y=y+1[/m] ⇒ [m] x-2y=1[/m]
Решаем систему двух уравнений:
[m]\left\{\begin {matrix}5x-8y=11\\x-2y=1\end {matrix}\right.[/m]
способом подстановки:
[m]\left\{\begin {matrix}5\cdot (2y+1)-8y=11\\x=2y+1\end {matrix}\right.[/m];
[m]\left\{\begin {matrix}10y+5-8y=11\\x=2y+1\end {matrix}\right.[/m];
[m]\left\{\begin {matrix}2y=6\\x=2y+1\end {matrix}\right.[/m];
[m]\left\{\begin {matrix}y=3\\x=2\cdot 3+1\end {matrix}\right.[/m];
[m]\left\{\begin {matrix}y=3\\x=7\end {matrix}\right.[/m]. Это ответ