$2$. Gọi cấp số nhân có số hạng đầu $u_1$, công bội $q$
$\Rightarrow u_2=u_1q;u_3=u_1q^2$
Theo bài ra ta có:
$\left\{\begin{array}{l} u_1+u_2+u_3=14\\u_1.u_2.u_3=64\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} u_1(1+q+q^2)=14\\u_1^3q^3=64\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} u_1(1+q+q^2)=14\\u_1q=4\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \dfrac{4}{q}(1+q+q^2)=14\\u_1=\dfrac{4}{q}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \dfrac{4}{q}+4+4q=14\\u_1=\dfrac{4}{q}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 4q^2-10q+4=0\\u_1=\dfrac{4}{q}\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} q=2;u_1=2\\q=\dfrac{1}{2};u_1=8\end{array} \right.$
