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Задача 57238 С y+2/x : C y+1/x : Cy/x = 7:7:5 ...

Условие

С y+2/x : C y+1/x : Cy/x = 7:7:5

103

Решение

[m]С^{y+2}_{x}=\frac{x!}{(x-(y+2))!\cdot (y+2)!}=\frac{x!}{(x-y-2)!\cdot (y+2)!}[/m]

[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}=7:7[/m]

⇒ [m]\frac{ x-y-1}{y+2}=\frac{7}{7}[/m] ⇒ [m]x-y-1=y+2)[/m] ⇒ [m] x=2y+3[/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}=7:5[/m]

⇒ [m]\frac{ x-y}{y+1}=\frac{7}{5}[/m] ⇒ [m]5(x-y)=7(y+1)[/m] ⇒ [m] x=\frac{12}{5}y+\frac{7}{5}[/m]

Решаем систему двух уравнений:
[m]\left\{\begin {matrix}x=2y+3\\ x=\frac{12}{5}y+\frac{7}{5}\end {matrix}\right.[/m] ; [m]\left\{\begin {matrix}x=2y+3\\2y+3=\frac{12}{5}y+\frac{7}{5}\end {matrix}\right.[/m] ; [m]\left\{\begin {matrix}x=2y+3\\\frac{2}{5}y=\frac{8}{5}\end {matrix}\right.[/m] [m]\left\{\begin {matrix}x=11\\y=4\end {matrix}\right.[/m]

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