Bài tập nâng cao · Bài 71
Bài tập nâng cao Hoán vị, chỉnh hợp và tổ hợp · Bài 71
Giải hệ phương trình $C_{x+1}^y:C_x^{y+1}:C_x^{y-1}=6:5:2$.
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Lời giải
Điều kiện: $\left\{\begin{array}{l}0\leq y\leq x+1\\0\leq y+1\leq x\\0\leq y-1\leq x\\x,y\in\mathbb{N}\end{array}\right.\Leftrightarrow\left\{\begin{array}{l}x,y\in\mathbb{N}\\y\geq1\\x\geq y+1\end{array}\right.$
Ta có:
$C_{x+1}^y:C_x^{y+1}:C_x^{y-1}=6:5:2\Leftrightarrow\frac{C_{x+1}^y}{6}=\frac{C_x^{y+1}}{5}=\frac{C_x^{y-1}}{2}\Leftrightarrow\left\{\begin{array}{l}\frac{C_{x+1}^y}{6}=\frac{C_x^{y+1}}{5}\\\frac{C_x^{y+1}}{5}=\frac{C_x^{y-1}}{2}\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}\frac{1}{6}\cdot\frac{(x+1)!}{y!(x+1-y)!}=\frac{1}{5}\cdot\frac{x!}{(y+1)!(x-y-1)!}\\\frac{1}{5}\cdot\frac{x!}{(y+1)!(x-y-1)!}=\frac{1}{2}\cdot\frac{x!}{(y-1)!(x-y+1)!}\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}5(x+1)(y+1)=6(x-y)(x-y+1)\\2(x-y)(x-y+1)=5y(y+1)\end{array}\right.\Leftrightarrow\left\{\begin{array}{l}5(x+1)(y+1)=3\cdot5y(y+1)\\2(x-y)(x-y+1)=5y(y+1)\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}x+1=3y\\2(x-y)(x-y+1)=5y(y+1)\end{array}\right.\Leftrightarrow\left\{\begin{array}{l}x=3y-1\\2(3y-1-y)(3y-1-y+1)=5y(y+1)\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}x=3y-1\\3y^2=9y\end{array}\right.\Leftrightarrow\left\{\begin{array}{l}x=3y-1\\y=3\end{array}\right.\Leftrightarrow\left\{\begin{array}{l}x=8\\y=3\end{array}\right.$
Vậy nghiệm của hệ là: $x=8$, $y=3$.
Ta có:
$C_{x+1}^y:C_x^{y+1}:C_x^{y-1}=6:5:2\Leftrightarrow\frac{C_{x+1}^y}{6}=\frac{C_x^{y+1}}{5}=\frac{C_x^{y-1}}{2}\Leftrightarrow\left\{\begin{array}{l}\frac{C_{x+1}^y}{6}=\frac{C_x^{y+1}}{5}\\\frac{C_x^{y+1}}{5}=\frac{C_x^{y-1}}{2}\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}\frac{1}{6}\cdot\frac{(x+1)!}{y!(x+1-y)!}=\frac{1}{5}\cdot\frac{x!}{(y+1)!(x-y-1)!}\\\frac{1}{5}\cdot\frac{x!}{(y+1)!(x-y-1)!}=\frac{1}{2}\cdot\frac{x!}{(y-1)!(x-y+1)!}\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}5(x+1)(y+1)=6(x-y)(x-y+1)\\2(x-y)(x-y+1)=5y(y+1)\end{array}\right.\Leftrightarrow\left\{\begin{array}{l}5(x+1)(y+1)=3\cdot5y(y+1)\\2(x-y)(x-y+1)=5y(y+1)\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}x+1=3y\\2(x-y)(x-y+1)=5y(y+1)\end{array}\right.\Leftrightarrow\left\{\begin{array}{l}x=3y-1\\2(3y-1-y)(3y-1-y+1)=5y(y+1)\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}x=3y-1\\3y^2=9y\end{array}\right.\Leftrightarrow\left\{\begin{array}{l}x=3y-1\\y=3\end{array}\right.\Leftrightarrow\left\{\begin{array}{l}x=8\\y=3\end{array}\right.$
Vậy nghiệm của hệ là: $x=8$, $y=3$.