Giải hệ phương trình sau:$\left\{ \begin{array}{l}C_{x + 1}^{y + 1} = C_{x + 1}^y\\3C_{x + 1}^{y + 1} = 5C_{x + 1}^{y - 1}\end{array} \right.$

Điều kiện $x,y \in \mathbb{N};\,x \ge y$

Ta có: $\left\{ \begin{array}{l}C_{x + 1}^{y + 1} = C_{x + 1}^y\\3C_{x + 1}^{y + 1} = 5C_{x + 1}^{y - 1}\end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l}\dfrac{{(x + 1)!}}{{(y + 1)!(x - y)!}} = \dfrac{{(x + 1)!}}{{y!(x - y + 1)!}}\\3\dfrac{{(x + 1)!}}{{(y + 1)!(x - y)!}} = 5\dfrac{{(x + 1)!}}{{(y - 1)!(x - y + 2)!}}\end{array} \right.$

$\Leftrightarrow \left\{ \begin{array}{l}\dfrac{1}{{y + 1}} = \dfrac{1}{{x - y + 1}}\\\dfrac{3}{{y(y + 1)}} = \dfrac{5}{{(x - y + 1)(x - y + 2)}}\end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l}x = 2y\\3(y + 1)(y + 2) = 5y(y + 1)\end{array} \right.$

$\Leftrightarrow \left\{ \begin{array}{l}x = 2y\\3y + 6 = 5y\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 6\\y = 3\end{array} \right.$ là nghiệm của hệ.