Gọi ba số đó là \({u_1},q{u_1},{q^2}{u_1}\). Ta có:
\(\left\{ \begin{array}{l}{u_1} + q{u_1} + {q^2}{u_1} = 27\\u_1^2 + {q^2}u_1^2 + {q^4}u_1^2 = 293\end{array} \right.\) \( \Rightarrow \left\{ \begin{array}{l}{u_1} = \dfrac{{27}}{{1 + q + {q^2}}}\\{\left( {\dfrac{{27}}{{1 + q + {q^2}}}} \right)^2}\left( {1 + {q^2} + {q^4}} \right) = 293\end{array} \right.\)
\( \Rightarrow 729\left( {1 + {q^2} + {q^4}} \right) = 293{\left( {1 + q + {q^2}} \right)^2}\)
\( \Leftrightarrow 729 + 729{q^2} + 729{q^4} = 293\left( {1 + {q^2} + {q^4} + 2q + 2{q^3} + 2{q^2}} \right)\)
\( \Leftrightarrow 436{q^4} - 586{q^3} - 150{q^2} - 586q + 436 = 0\)
\(\begin{array}{l} \Leftrightarrow 436{q^2} + \dfrac{{436}}{{{q^2}}} - \left( {586q + \dfrac{{586}}{q}} \right) - 150 = 0\\ \Leftrightarrow 436\left( {{q^2} + \dfrac{1}{{{q^2}}}} \right) - 586\left( {q + \dfrac{1}{q}} \right) - 150 = 0\end{array}\)
Đặt \(t = q + \dfrac{1}{q} \Rightarrow {t^2} = {q^2} + \dfrac{1}{{{q^2}}} + 2 \Rightarrow {q^2} + \dfrac{1}{{{q^2}}} = {t^2} - 2\)
\(\begin{array}{l} \Rightarrow 436\left( {{t^2} - 2} \right) - 586t - 150 = 0\\ \Leftrightarrow 436{t^2} - 586t - 1022 = 0 \Leftrightarrow \left[ \begin{array}{l}t = \dfrac{{511}}{{218}}\\t = - 1\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}q + \dfrac{1}{q} = \dfrac{{511}}{{218}}\\q + \dfrac{1}{q} = - 1\end{array} \right. \Rightarrow q = ...\end{array}\)
