Đáp án:
a) $(q;u_1)= (2;9)$
b) $(q ; u_1 )=(3; \dfrac{1}{3}),(\dfrac{1}{3};243)$
Giải thích các bước giải:
a, Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
{u_4} - {u_2} = 54\\
{u_5} - {u_3} = 108
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{u_1}.{q^3} - {u_1}.q = 54\\
{u_1}.{q^4} - {u_1}.{q^2} = 108
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1}.q.\left( {{q^2} - 1} \right) = 54\\
{u_1}.{q^2}\left( {{q^2} - 1} \right) = 108
\end{array} \right. \Rightarrow \frac{q}{{{q^2}}} = \frac{{54}}{{108}} \Rightarrow q = 2 \Rightarrow {u_1} = 9
\end{array}\)
b, \(\begin{array}{l}
\left\{ \begin{array}{l}
{u_2} + {u_4} + {u_6} = 91\\
{u_3} + {u_5} = 30
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{u_1}.q + {u_1}.{q^3} + {u_1}.{q^5} = 91\\
{u_1}.{q^2} + {u_1}.{q^4} = 30
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1}.q.\left( {1 + {q^2} + {q^4}} \right) = 91\\
{u_1}.q\left( {q + {q^3}} \right) = 30
\end{array} \right. \Rightarrow \frac{{1 + {q^2} + {q^4}}}{{q + {q^3}}} = \frac{{91}}{{30}}\\
\Leftrightarrow 30{q^4} - 91{q^3} + 30{q^2} - 91q + 30 = 0\\
\Leftrightarrow \left( {30{q^4} - 90{q^3}} \right) - \left( {{q^3} - 3{q^2}} \right) + \left( {27{q^2} - 81q} \right) - \left( {10q - 30} \right) = 0\\
\Leftrightarrow \left( {q - 3} \right)\left( {30{q^3} - {q^2} + 27q - 10} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
q = 3 \Rightarrow {u_1} = \frac{1}{3}\\
q = \frac{1}{3} \Rightarrow u_1=243
\end{array} \right.
\end{array}\)
