5.034 (б). Найти \(x\) и \(y\), если:
$$C^{y-1}_x:(C^y_{x-2}+C^{y-2}_{x-2}+2C^{y-1}_{x-2}):C^{y+1}_x=3:5:5.$$
Решение.
\[C^{y-1}_x:(C^y_{x-2}+C^{y-2}_{x-2}+2C^{y-1}_{x-2}):C^{y+1}_x=3:5:5\]
\(C_n^m+C_n^{m+1}=C_{n+1}^{m+1}\)
\begin{multline}
C^y_{x-2}+C^{y-2}_{x-2}+2C^{y-1}_{x-2}=C^y_{x-2}+C^{y-2}_{x-2}+C^{y-1}_{x-2}+C^{y-1}_{x-2}=\\=(C^{y-1}_{x-2}+C^y_{x-2})+(C^{y-2}_{x-2}+C^{y-1}_{x-2})=C^y_{x-1}+C^{y-1}_{x-1}=C^{y-1}_{x-1}+C^y_{x-1}=C^y_x
\end{multline}
\(C^{y-1}_x:C^y_x:C^{y+1}_x=3:5:5\)
$$\left\{\begin{array}{rcl}C^{y-1}_x:C^y_x=3:5\\C^y_x:C^{y+1}_x=5:5\end{array}\right.$$
\(C_n^m = \frac{n!}{(n - m)! m!}\), где \(m \leq n; C_n^0 = 1;\)
\(C^{y-1}_x=\frac{x!}{(x-(y-1))!\cdot(y-1)!}=\frac{x!}{(x-y+1)!(y-1)!}\)
\(C^y_x=\frac{x!}{(x-y)!\cdot y!}\)
\(C^{y+1}_x=\frac{x!}{(x-(y+1))!\cdot(y+1)!}=\frac{x!}{(x-y-1)!(y+1)!}\)
$$\left\{\begin{array}{rcl}\frac{x!}{(x-y+1)!(y-1)!}:\frac{x!}{(x-y)!\cdot y!}=\frac{3}{5} \\ \frac{x!}{(x-y)!\cdot y!}:\frac{x!}{(x-y-1)!(y+1)!} =\frac{5}{5}\end{array}\right.$$
$$\left\{\begin{array}{rcl}\frac{x!}{(x-y+1)!(y-1)!}\cdot\frac{(x-y)!\cdot y!}{x!}=\frac{3}{5} \\ \frac{x!}{(x-y)!\cdot y!}\cdot\frac{(x-y-1)!(y+1)!}{x!} =1\end{array}\right.$$
$$\left\{\begin{array}{rcl}\frac{x!}{(x-y+1)(x-y)!(y-1)!}\cdot\frac{(x-y)!\cdot y\cdot(y-1)!}{x!}=\frac{3}{5} \\ \frac{x!}{(x-y)(x-y-1)!\cdot y!}\cdot\frac{(x-y-1)!(y+1)\cdot y!}{x!} =1\end{array}\right.$$
$$\left\{\begin{array}{rcl}\frac{y}{x-y+1}=\frac{3}{5} \\ \frac{y+1}{x-y}=1\end{array}\right.$$
Значит, \(x-y=y+1 \Rightarrow x=2y+1\).
$$\left\{\begin{array}{rcl}\frac{y}{x-y+1}=\frac{3}{5} \\ x=2y+1\end{array}\right.$$
$$\left\{\begin{array}{rcl}\frac{y}{2y+1-y+1}=\frac{3}{5} \\ x=2y+1\end{array}\right.$$
Решим уравнение \(\frac{y}{2y+1-y+1}=\frac{3}{5}\) и найдём чему равна переменная \(y\):
\(\frac{y}{2y+1-y+1}=\frac{3}{5}\)
\(\frac{y}{y+2}=\frac{3}{5}\)
\(3\cdot(y+2)=5y\)
\(3y+6=5y\)
\(5y-3y=6\)
\(2y=6\)
\(y=3\).
Так как \(x=2y+1\), то \(x=2\cdot3+1 \Rightarrow x=6+1=7\). Значит, \(x=7\).
Проверка.
При \(x=7\) и \(y=3\):
\(C^{y-1}_x=C^{3-1}_7=C^2_7\);
\(C^y_{x-2}=C^3_{7-2}=C^3_5\);
\(C^{y-2}_{x-2}=C^{3-2}_{7-2}=C^1_5\);
\(2C^{y-1}_{x-2}=2C^{3-1}_{7-2}=2C^2_5\);
\(C^{y+1}_x=C^{3+1}_7=C^4_7\).
Значит, отношение \(C^2_7:(C^3_5+C^1_5+2C^2_5):C^4_7\) должно равняться отношению \(3:5:5\).
\(C_7^2=\frac{7!}{(7-2)!\cdot2!}=\frac{7\cdot6\cdot5!}{5!\cdot2}=\frac{7\cdot6}{2}=7\cdot3=21\)
\(C_5^3=\frac{5!}{(5-3)!\cdot3!}=\frac{5\cdot4\cdot3!}{2!\cdot3!}=\frac{5\cdot4}{2}=5\cdot2=10\)
\(C_5^1=\frac{5!}{(5-1)!\cdot1!}=\frac{5\cdot4!}{4!\cdot1}=\frac{5}{1}=5\)
\(2C_5^2=2\cdot\frac{5!}{(5-2)!\cdot2!}=2\cdot\frac{5\cdot4\cdot3!}{3!\cdot2!}=2\cdot\frac{5\cdot4}{2}=5\cdot4=20\)
\(C_7^4=\frac{7!}{(7-4)!\cdot4!}=\frac{7\cdot6\cdot5\cdot4!}{3!\cdot4!}=\frac{7\cdot6\cdot5}{3\cdot2}=\frac{7\cdot6\cdot5}{6}=7\cdot5=35\)
\(C^{y-1}_x:(C^y_{x-2}+C^{y-2}_{x-2}+2C^{y-1}_{x-2}):C^{y+1}_x=C^2_7:(C^3_5+C^1_5+2C^2_5):C^4_7=21:(10+5+20):35=21:35:35=3:5:5\)
\(C^{y-1}_x:(C^y_{x-2}+C^{y-2}_{x-2}+2C^{y-1}_{x-2}):C^{y+1}_x=3:5:5\)
Значит, \(x=7\) и \(y=3\).
Ответ: \(x=7\); \(y=3\).