Bài tập nâng cao · Bài 71
Bài tập nâng cao Hoán vị, chỉnh hợp và tổ hợp · Bài 71
Giải hệ phương trình sau: $C_{x+1}^y:C_x^{y+1}:C_x^{y-1}=6:5:2$.
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Lời giải
Điều kiện: $\left\{\begin{array}{l}0\le y\le x+1\\0\le y+1\le x\\0\le y-1\le x\\x,y\in\mathbb N\end{array}\right.\Leftrightarrow\left\{\begin{array}{l}x,y\in\mathbb N\\y\ge1\\x\ge y+1\end{array}\right.$.
Ta có:
$C_{x+1}^y:C_x^{y+1}:C_x^{y-1}=6:5:2\Leftrightarrow\dfrac{C_{x+1}^y}{6}=\dfrac{C_x^{y+1}}5=\dfrac{C_x^{y-1}}2\Leftrightarrow\left\{\begin{array}{l}\dfrac{C_{x+1}^y}{6}=\dfrac{C_x^{y+1}}5\\\dfrac{C_x^{y+1}}5=\dfrac{C_x^{y-1}}2\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}\dfrac16\cdot\dfrac{(x+1)!}{y!(x+1-y)!}=\dfrac15\cdot\dfrac{x!}{(y+1)!(x-y-1)!}\\\dfrac15\cdot\dfrac{x!}{(y+1)!(x-y-1)!}=\dfrac12\cdot\dfrac{x!}{(y-1)!(x-y+1)!}\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}5(x+1)(y+1)=6(x-y)(x-y+1)\\2(x-y)(x-y+1)=5y(y+1)\end{array}\right.\Leftrightarrow\left\{\begin{array}{l}5(x+1)(y+1)=3\cdot5y(y+1)\\2(x-y)(x-y+1)=5y(y+1)\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}x+1=3y\\2(x-y)(x-y+1)=5y(y+1)\end{array}\right.\Leftrightarrow\left\{\begin{array}{l}x=3y-1\\2(3y-1-y)(3y-1-y+1)=5y(y+1)\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}x=3y-1\\3y^2=9y\end{array}\right.\Leftrightarrow\left\{\begin{array}{l}x=3y-1\\y=3\end{array}\right.\Leftrightarrow\left\{\begin{array}{l}x=8\\y=3\end{array}\right.$.
Vậy nghiệm của hệ là: $x=8$, $y=3$.
Ta có:
$C_{x+1}^y:C_x^{y+1}:C_x^{y-1}=6:5:2\Leftrightarrow\dfrac{C_{x+1}^y}{6}=\dfrac{C_x^{y+1}}5=\dfrac{C_x^{y-1}}2\Leftrightarrow\left\{\begin{array}{l}\dfrac{C_{x+1}^y}{6}=\dfrac{C_x^{y+1}}5\\\dfrac{C_x^{y+1}}5=\dfrac{C_x^{y-1}}2\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}\dfrac16\cdot\dfrac{(x+1)!}{y!(x+1-y)!}=\dfrac15\cdot\dfrac{x!}{(y+1)!(x-y-1)!}\\\dfrac15\cdot\dfrac{x!}{(y+1)!(x-y-1)!}=\dfrac12\cdot\dfrac{x!}{(y-1)!(x-y+1)!}\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}5(x+1)(y+1)=6(x-y)(x-y+1)\\2(x-y)(x-y+1)=5y(y+1)\end{array}\right.\Leftrightarrow\left\{\begin{array}{l}5(x+1)(y+1)=3\cdot5y(y+1)\\2(x-y)(x-y+1)=5y(y+1)\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}x+1=3y\\2(x-y)(x-y+1)=5y(y+1)\end{array}\right.\Leftrightarrow\left\{\begin{array}{l}x=3y-1\\2(3y-1-y)(3y-1-y+1)=5y(y+1)\end{array}\right.$
$\Leftrightarrow\left\{\begin{array}{l}x=3y-1\\3y^2=9y\end{array}\right.\Leftrightarrow\left\{\begin{array}{l}x=3y-1\\y=3\end{array}\right.\Leftrightarrow\left\{\begin{array}{l}x=8\\y=3\end{array}\right.$.
Vậy nghiệm của hệ là: $x=8$, $y=3$.