Bài tập 3.30 trang 131 SBT Toán 11

a) Ta có:

\(\begin{array}{l}
\left\{ \begin{array}{l}
{u_5} - {u_1} = 15\\
{u_4} - {u_2} = 6
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{u_1}{q^4} - {u_1} = 15\\
{u_1}{q^3} - {u_1}q = 6
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{u_1}\left( {{q^2} + 1} \right)\left( {{q^2} - 1} \right) = 15\\
{u_1}q\left( {{q^2} - 1} \right) = 6
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left( {{q^2} + 1} \right).\frac{6}{q} = 15\\
{u_1}\left( {{q^2} - 1} \right) = \frac{6}{q}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
6{q^2} - 15q + 6 = 0\\
{u_1}\left( {{q^2} - 1} \right) = \frac{6}{q}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
q = 2\\
{u_1} = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
q = \frac{1}{2}\\
{u_1} =  - 16
\end{array} \right.
\end{array} \right.
\end{array}\)

b) \(\left\{ \begin{array}{l}
{u_2} - {u_4} + {u_5} = 10\\
{u_3} - {u_5} + {u_6} = 20
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{u_1}q - {u_1}{q^3} + {u_1}{q^4} = 10\\
{u_1}{q^2} - {u_1}{q^4} + {u_1}{q^5} = 20
\end{array} \right.\)

\(\begin{array}{l}
 \Leftrightarrow \left\{ \begin{array}{l}
{u_1}q\left( {1 - {q^2} + {q^3}} \right) = 10\\
{u_1}{q^2}\left( {1 - {q^2} + {q^3}} \right) = 20
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{u_1}q\left( {1 - {q^2} + {q^3}} \right) = 10\\
q.10 = 20
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
q = 2\\
{u_1} = 1
\end{array} \right.
\end{array}\)

-- Mod Toán 11 HỌC247