Bài toán 1: Tìm số hạng tổng quát của dãy số $\left \{ x_{n} \right \}$ sau: $\left\{\begin{matrix} x_{1}=\alpha >0\\ x_{n+1}=\frac{x_{n}^{3}+12x_{n}}{3x_{n}^{2}+4} \end{matrix}\right.$ $\forall n=1,2,...$
Ta có:
$x_{n+1}- 2 = \dfrac{x_{n}^{3}+12x_n}{3x_n^2+4}-2=\dfrac{x_{n}^{3}-6x_n^2+12x_n-8}{3x_n^2+4}=\dfrac{\left (x_n-2 \right )^3}{3x_n^2+4}$
$x_{n+1} + 2 = \dfrac{\left (x_n+2 \right )^3}{3x_n^2+4}$
Xét hàm số $f(x)=\dfrac{x-2}{x+2}$, ta thấy $f\left ( x_{n+1} \right ) = \dfrac{x_{n+1}-2}{x_{n+1}+2} = \dfrac{\left (x_n-2 \right )^3}{\left (x_n+2 \right )^3} = f^3\left ( x_{n} \right )$
$\Rightarrow f\left ( x_{n} \right ) = f^{3^{n-1}}\left ( x_{1} \right )=f^{3^{n-1}}\left ( \alpha \right )$
Đặt $f^{3^{n-1}}\left ( \alpha \right ) = \beta$ thì $\dfrac{x_n-2}{x_n+2}= \beta \Leftrightarrow x_n = \dfrac{2+2\beta}{1-\beta}$
Vậy $x_n = \dfrac{2+2\beta}{1-\beta}$ với $\beta = \left (\dfrac{\alpha -2}{\alpha+2} \right )^{3^{n-1}}$