Đáp án:
\(\left\{ \begin{array}{l}
q = \pm \dfrac{1}{2}\\
{u_1} = 28
\end{array} \right.\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
{u_1} + {u_2} + {u_3} + {u_4} + {u_5} = 49.\left( {\dfrac{1}{{{u_1}}} + \dfrac{1}{{{u_2}}} + \dfrac{1}{{{u_3}}} + \dfrac{1}{{{u_4}}} + \dfrac{1}{{{u_5}}}} \right)\\
{u_1} + {u_3} = 35
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1}\left( {1 + q + {q^2} + {q^3} + {q^4}} \right) = \dfrac{{49}}{{{u_1}}}\left( {1 + \dfrac{1}{q} + \dfrac{1}{{{q^2}}} + \dfrac{1}{{{q^3}}} + \dfrac{1}{{{q^4}}}} \right)\\
{u_1} + {u_1}.{q^2} = 35
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1}.\dfrac{{{q^5} - 1}}{{q - 1}} = \dfrac{{49}}{{{u_1}}}.\dfrac{{{{\left( {\frac{1}{q}} \right)}^5} - 1}}{{\frac{1}{q} - 1}}\\
{u_1} = \dfrac{{35}}{{1 + {q^2}}}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{35}}{{1 + {q^2}}}.\dfrac{{{q^5} - 1}}{{q - 1}} = \dfrac{{49}}{{\frac{{35}}{{1 + {q^2}}}}}.\dfrac{{\frac{{1 - {q^5}}}{{{q^5}}}}}{{\frac{{1 - q}}{q}}}\\
{u_1} = \dfrac{{35}}{{1 + {q^2}}}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{35}}{{1 + {q^2}}}.\dfrac{{{q^5} - 1}}{{q - 1}} = \dfrac{{49\left( {1 + {q^2}} \right)}}{{35}}.\dfrac{{1 - {q^5}}}{{\left( {1 - q} \right).{q^4}}}\\
{u_1} = \dfrac{{35}}{{1 + {q^2}}}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{35}}{{1 + {q^2}}} = \dfrac{{49\left( {1 + {q^2}} \right)}}{{35{q^4}}}\\
{u_1} = \dfrac{{35}}{{1 + {q^2}}}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
1225{q^4} = 49\left( {1 + 2{q^2} + {q^4}} \right)\\
{u_1} = \dfrac{{35}}{{1 + {q^2}}}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
1176{q^4} - 98{q^2} - 49 = 0\\
{u_1} = \dfrac{{35}}{{1 + {q^2}}}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
{q^2} = - \dfrac{1}{6}\left( l \right)\\
{q^2} = \dfrac{1}{4}
\end{array} \right.\\
{u_1} = \dfrac{{35}}{{1 + {q^2}}}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
q = \pm \dfrac{1}{2}\\
{u_1} = 28
\end{array} \right.
\end{array}\)
