Tìm cặp số (x;y) thỏa mãn $\Large \dfrac{C_{x+1}^{y}}{6}=\dfrac{C_{x}^

Tìm cặp số (x;y) thỏa mãn $\Large \dfrac{C_{x+1}^{y}}{6}=\dfrac{C_{x}^{y+1}}{5}=\dfrac{C_{z}^{y-1}}{2}$

• A. $\Large (x ; y)=(8 ; 3)$
• B. $\Large (x ; y)=(3 ; 8)$
• C. $\Large (x ; y)=(-1 ; 0)$
• D. $\Large (x ; y)=(-1 ; 0),(x ; y)=(8 ; 3)$

Điều kiện: $\Large x \geq y+1$ và $\Large $\Large x, y \in \mathbb N$

$\Large \dfrac{C_{x+1}^{y}}{6}=\dfrac{C_{x}^{y+1}}{5} \Leftrightarrow 5 \cdot C_{x+1}^{y}=6 . C_{x}^{y+1}$ $\Large \Leftrightarrow \dfrac{5(x+1) !}{y !(x+1-y) !}=\dfrac{6 x !}{(y+1) !(x-y-1) !}$

$\Large \Leftrightarrow \dfrac{5(x+1)}{(x-y)(x-y+1)}=\dfrac{6}{(y+1)}$ $\Large \Leftrightarrow 5(y+1)(x+1)=6(x-y)(x-y+1)$ . (1)

$\Large \dfrac{C_{x}^{y+1}}{5}=\dfrac{C_{x}^{y-1}}{2} \Leftrightarrow 2 . C_{x}^{y+1}=5 . C_{x}^{y-1}$ $\Large \Leftrightarrow \dfrac{x !}{5 \cdot(y+1) !(x-y-1) !}=\dfrac{x !}{2 \cdot(y-1) ! \cdot(x-y+1) !}$

$\Large \Leftrightarrow \dfrac{1}{5 . y(y+1)}=\dfrac{1}{2 \cdot(x-y)(x-y+1)}$ 

$\Large \Leftrightarrow 5 . y(y+1)=2 \cdot(x-y)(x-y+1)$ $\Large \Leftrightarrow 15 . y(y+1)=6 .(x-y)(x-y+1)$. (2)

Từ (1) và (2) suy ra:

$\Large \Leftrightarrow 5(y+1)(x+1)=15 . y(y+1) \Leftrightarrow x+1=3 y$

Thay vào (1), ta được

$\Large \Leftrightarrow 15(y+1) y=6(2 y-1) 2 y \Leftrightarrow 3 y^{2}-9 y=0$ $\Large \Leftrightarrow\left[\begin{array}{l}
y=0 \rightarrow x=-1(\text { loai }) \\
y=3 \rightarrow x=8(\text { nhan })
\end{array}\right.$

Chọn A