Bài tập 3.20 trang 124 SBT Toán 11

a) \(\left\{ \begin{array}{l}
{u_1} + 2{u_5} = 0\\
{S_4} = 14
\end{array} \right.\)

\(\begin{array}{l}
 \Leftrightarrow \left\{ \begin{array}{l}
{u_1} + 2.\left( {{u_1} + 4d} \right) = 0\\
4{u_1} + \frac{{4.3}}{2}.d = 14
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
3{u_1} + 8d = 0\\
4{u_1} + 6d = 14
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{u_1} = 8\\
d =  - 3
\end{array} \right.
\end{array}\)

b) \(\left\{ \begin{array}{l}
{u_4} = 10\\
{u_7} = 19
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{u_1} + 3d = 10\\
{u_1} + 6d = 19
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{u_1} = 1\\
d = 3
\end{array} \right.\)

c) \(\left\{ \begin{array}{l}
{u_1} + {u_5} - {u_3} = 10\\
{u_1} + {u_6} = 7
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{u_1} + 2d = 10\\
2{u_1} + 5d = 7
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{u_1} = 36\\
d =  - 13
\end{array} \right.\)

d) \(\left\{ \begin{array}{l}
{u_7} - {u_3} = 8\\
{u_2}.{u_1} = 75
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
4d = 8\\
u_1^2 + {u_1}.d = 75
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
d = 2\\
\left[ \begin{array}{l}
{u_1} = 3\\
{u_1} =  - 17
\end{array} \right.
\end{array} \right.\)

-- Mod Toán 11 HỌC247