Задача 57075 Из данной пропорции найти Х и Y...

6004635c960c3c35933dfc4c

17.01.2021 19:32:56

Из данной пропорции найти Х и Y

5f3ea7e3faf909182968ddd9

17.01.2021 20:00:25

[m]С^{y+1}_{x+1}=\frac{(x+1)!}{(x+1-(y+1))!\cdot (y+1)!}=\frac{(x+1)!}{(x+1-y-1))!\cdot (y+1)!}=\frac{(x+1)!}{(x-y))!\cdot (y+1)!}[/m]

[m]С^{y}_{x+1}=\frac{(x+1)!}{(x+1–y)!\cdot y!}[/m]

[m]С^{y–1}_{x+1}=\frac{(x+1)!}{((x+1)-(y-1))!\cdot (y-1)!}[/m]



[m]С^{y+1}_{x+1}:С^{y}_{x+1}=\frac{(x+1)!}{(x-y))!\cdot (y+1)!}:\frac{(x+1)!}{(x+1–y)!\cdot y!}=\frac{(x+1)!}{(x-y))!\cdot (y+1)!}\cdot \frac{(x+1–y)!\cdot y!}{(x+1)!}=\frac{(x+1–y)!\cdot y!}{(x-y))!\cdot (y+1)!}=\frac{(x-y)!\cdot (x–y+1)\cdot y!}{(x-y))!\cdot y!\cdot (y+1)}=\frac{x–y+1}{y+1}[/m]

[m]С^{y}_{x+1}:С^{y–1}_{x+1}=\frac{(x+1)!}{(x+1–y)!\cdot y!}:\frac{(x+1)!}{((x+1)-(y-1))!\cdot (y-1)!}=\frac{(x+1)!}{(x+1–y)!\cdot y!}\cdot \frac{((x+1)-y+1))!\cdot (y-1)!}{(x+1)!}=\frac{x-y+2}{y}[/m]

По условию:
[m]С^{y+1}_{x+1}:С^{y}_{x+1}=14:7[/m]

⇒ [m]\frac{x–y+1}{y+1}=\frac{14}{7}[/m] ⇒ [m]7\cdot (x-y+1)=14(y+1)[/m] ⇒ [m] x-y+1=2y+2[/m]

[m]С^{y}_{x+1}:С^{y–1}_{x+1}=7:2[/m]

⇒ [m]\frac{x-y+2}{y}=\frac{7}{2}[/m] ⇒ [m]2\cdot (x-y+2)=7y[/m] ⇒ [m]2x-2y+4=7y[/m]

Решаем систему двух уравнений:
[m]\left\{\begin {matrix}x-y+1=2y+2\\2x-2y+4=7y\end {matrix}\right.[/m][m]\left\{\begin {matrix}x=3y+1\\9 y=2x+4\end {matrix}\right.[/m][m]\left\{\begin {matrix}x=3y+1\\9 y=2(3y+1)+4\end {matrix}\right.[/m]

[m]\left\{\begin {matrix}x=3y+1\\3 y=6\end {matrix}\right.[/m]

[m]\left\{\begin {matrix}x=3\cdot 2+1\\ y=2\end {matrix}\right.[/m]

О т в е т. [m]x=7;y=2[/m]