a) Ta có hệ \(\left\{ \begin{array}{l}{u_1}{q^4} - {u_1} = 15\\{u_1}{q^3} - {u_1}q = 6\end{array} \right.\) hay \(\left\{ \begin{array}{l}{u_1}\left( {{q^4} - 1} \right) = 15\\{u_1}\left( {{q^3} - q} \right) = 6.\end{array} \right.{\rm{ }}\left( 1 \right)\)
Do (1) nên \(q \ne \pm 1,\) suy ra \(\dfrac{{15}}{6} = \dfrac{{{q^4} - 1}}{{q\left( {{q^2} - 1} \right)}} = \dfrac{{{q^2} + 1}}{q}.\)
Biến đổi về phương trình \(2{q^2} - 5q + 2 = 0.\)
Giải ra được \(q = 2\) và \(q = \dfrac{1}{2}.\)
Nếu \(q = 2\) thì \({u_1} = 1.\)
Nếu \(q = \dfrac{1}{2}\) thì \({u_1} = - 16.\)
b) Ta có: \(\left\{ \begin{array}{l}{u_1}q - {u_1}{q^3} + {u_1}{q^4} = 10\\{u_1}{q^2} - {u_1}{q^4} + {u_1}{q^5} = 20\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}{u_1}q - {u_1}{q^3} + {u_1}{q^4} = 10\\q\left( {{u_1}q - {u_1}{q^3} + {u_1}{q^4}} \right) = 20\end{array} \right.\)
\( \Leftrightarrow \left\{ \begin{array}{l}q = 2\\2{u_1} - 8{u_1} + 16{u_1} = 10\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}q = 2\\{u_1} = 1\end{array} \right.\)
Vậy \({u_1} = 1,q = 2.\)
