Giải thích các bước giải:
Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
{u_1} + {u_3} = 2525\\
{u_2} - {u_4} = \dfrac{{1515}}{2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1}\left( {1 + {q^2}} \right) = 2525\\
{u_1}\left( {q - {q^3}} \right) = \dfrac{{1515}}{2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1}\left( {1 + {q^2}} \right) = 2525\\
\dfrac{{1 + {q^2}}}{{q - {q^3}}} = \dfrac{{10}}{3}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1}\left( {1 + {q^2}} \right) = 2525\\
10{q^3} + 3{q^2} - 10q + 3 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1}\left( {1 + {q^2}} \right) = 2525\\
\left( {2q - 1} \right)\left( {5{q^2} + 4q - 3} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1} = \dfrac{{2525}}{{1 + {q^2}}}\\
\left[ \begin{array}{l}
q = \dfrac{1}{2}\\
q = \dfrac{{ - 2 + \sqrt {19} }}{5}\\
q = \dfrac{{ - 2 - \sqrt {19} }}{5}
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
q = \dfrac{1}{2};{u_1} = 2020\\
q = \dfrac{{ - 2 + \sqrt {19} }}{5};{u_1} \approx 2065,31\\
q = \dfrac{{ - 2 - \sqrt {19} }}{5};{u_1} \approx 964,69
\end{array} \right.
\end{array}$
Vậy $\left( {{u_1};q} \right) \in \left\{ {\left( {2020;\dfrac{1}{2}} \right);\left( {2065,31;\dfrac{{ - 2 + \sqrt {19} }}{5}} \right);\left( {964,69;\dfrac{{ - 2 - \sqrt {19} }}{5}} \right)} \right\}$
