Задача 61779 Из данной пропорции найти x и y ...
Условие
Решение
[m]С^{y+1}_{x+1}=\frac{(x+1)!}{(y+1)!(x+1-(y+1))!}[/m]
[m]С^{y}_{x+1}=\frac{(x+1)!}{y!(x+1-y)!}[/m]
[m]С^{y+2}_{x+y}:С^{y+1}_{x+1}:С^{y}_{x+1}=3:4:3[/m]
⇒
[m]С^{y+2}_{x+y}:С^{y+1}_{x+1}=3:4[/m]⇒[m]\frac{(x+y)!}{(y+2)!(x+y-(y+2))!}:\frac{(x+1)!}{(y+1)!(x+1-(y+1))!}=3:4[/m]⇒[m]\frac{(x+y)!}{(y+2)!(x-2)!}\cdot \frac{(y+1)!(x-y)!}{(x+1)!}=3:4[/m]
[m]С^{y+1}_{x+1}:С^{y}_{x+1}=4:3[/m]⇒ [m]\frac{(x+1)!}{(y+1)!(x+1-(y+1))!}:\frac{(x+1)!}{y!(x+1-y)!}=4:3[/m]⇒[m]\frac{(x+1)!}{(y+1)!(x-y)!}\cdot \frac{y!(x+1-y)!}{(x+1)!}=4:3[/m]
[m]С^{y+2}_{x+y}:С^{y}_{x+1}=3:3[/m]⇒[m]\frac{(x+y)!}{(y+2)!(x+y-(y+2))!}:\frac{(x+1)!}{y!(x+1-y)!}=3:3[/m]⇒[m]\frac{(x+y)!}{(y+2)!(x-2)!}\cdot \frac{y!(x+1-y)!}{(x+1)!}=3:3[/m]
Получаем систему:
[m]\left\{\begin {matrix}\frac{(x+y)!}{(y+2)!(x-2)!}\cdot \frac{(y+1)!(x-y)!}{(x+1)!}=3:4\\\frac{(x+1)!}{(y+1)!(x-y)!}\cdot \frac{y!(x+1-y)!}{(x+1)!}=4:3\\\frac{(x+y)!}{(y+2)!(x-2)!}\cdot \frac{y!(x+1-y)!}{(x+1)!}=3:3\end {matrix}\right.[/m]
[m](y+1)!=y!\cdot (y+1)[/m]
[m](y+2)!=y!\cdot (y+1)(y+2)[/m]
[m](x-y+1)!=(x-y)!\cdot(x-y+1)[/m]
[m]\left\{\begin {matrix}\frac{(x+y)!}{(y+1)!\cdot (y+2)\cdot (x-2)!}\cdot \frac{(y+1)!(x-y)!}{(x+1)!}=3:4\\\frac{1}{(y+1)!(x-y)!}\cdot \frac{y!(x+1-y)!}{1}=4:3\\\frac{(x+y)!}{y!\cdot (y+1)\cdot (y+2)(x-2)!}\cdot \frac{y!(x+1-y)!}{(x+1)!}=1\end {matrix}\right.[/m]
[m](x-y+1)!=(x-y)!\cdot(x-y+1)[/m]
[m]\left\{\begin {matrix}\frac{(x+y)!}{ (y+2)\cdot (x-2)!}\cdot \frac{(x-y)!}{(x+1)!}=3:4\\\frac{1}{(y+1)(x-y)!}\cdot \frac{(x-y)!\cdot (x+1-y)}{1}=4:3\\\frac{(x+y)!}{ (y+1)\cdot (y+2)\cdot (x-2)!}\cdot \frac{(x+1-y)!}{(x+1)!}=1\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}\frac{(x+y)!}{ (y+2)\cdot (x-2)!}\cdot \frac{(x-y)!}{(x+1)!}=3:4\\\frac{1}{(y+1)}\cdot \frac{ (x+1-y)}{1}=4:3\\\frac{(x+y)!}{ (y+1)\cdot (y+2)\cdot (x-2)!}\cdot \frac{(x+1-y)!}{(x+1)!}=1\end {matrix}\right.[/m]
[m]\left\{\begin {matrix}4\cdot (x+y)!\cdot (x-y)!=3\cdot (y+2)\cdot (x-2)!\cdot (x+1)!\\3(x+1-y)=4(y+1)\\(x+y)!\cdot (x+1-y)!= (y+1)\cdot (y+2)\cdot (x-2)!\cdot(x+1)!\end {matrix}\right.[/m]