Giải bài 9 trang 107 – SGK môn Đại số và Giải tích 11
Tìm số hạng đầu \(u_1\) và công bội \(q\) của các cấp số nhân (\(u_n\)), biết:
a) \(\left\{ \begin{aligned} & {{u}_{6}}=192 \\ & {{u}_{7}}=384 \\ \end{aligned} \right. \)
b) \(\left\{ \begin{aligned} & {{u}_{4}}-{{u}_{2}}=72 \\ & {{u}_{5}}-{{u}_{3}}=144 \\ \end{aligned} \right. \)
c) \( \left\{ \begin{aligned} & {{u}_{2}}+{{u}_{5}}-{{u}_{4}}=10 \\ & {{u}_{3}}+{{u}_{6}}-{{u}_{5}}=20 \\ \end{aligned} \right. \)
Hướng dẫn:
Áp dụng: \(u_n=u_1.q^{n-1}\)
a)
\(\left\{ \begin{aligned} & {{u}_{6}}=192 \\ & {{u}_{7}}=384 \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & {{u}_{1}}{{q}^{5}}=192 \\ & {{u}_{1}}{{q}^{6}}=384 \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & q=2 \\ & {{u}_{1}}.2^6=384 \\ \end{aligned} \right. \Leftrightarrow \left\{ \begin{aligned} & q=2 \\ & u_1=6 \\ \end{aligned} \right. \)
b)
\(\begin{aligned} & \left\{ \begin{aligned} & {{u}_{4}}-{{u}_{2}}=72 \\ & {{u}_{5}}-{{u}_{3}}=144 \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & {{u}_{1}}{{q}^{3}}-{{u}_{1}}q=72 \\ & {{u}_{1}}{{q}^{4}}-{{u}_{1}}{{q}^{2}}=144 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & {{u}_{1}}q\left( {{q}^{2}}-1 \right)=72 \\ & {{u}_{1}}{{q}^{2}}\left( {{q}^{2}}-1 \right)=144 \\ \end{aligned} \right.\\&\Leftrightarrow \left\{ \begin{aligned} & q=2 \\ & {{u}_{1}}.2^2(2^2-1)=144 \\ \end{aligned} \right. \\&\Leftrightarrow \left\{ \begin{aligned} & q=2 \\ &12 {{u}_{1}}=144 \\ \end{aligned} \right.\\&\Leftrightarrow \left\{ \begin{aligned} & q=2 \\ & {{u}_{1}}=12 \\ \end{aligned} \right. \\ \end{aligned} \)
c)
\(\begin{aligned} & \left\{ \begin{aligned} & {{u}_{2}}+{{u}_{5}}-{{u}_{4}}=10 \\ & {{u}_{3}}+{{u}_{6}}-{{u}_{5}}=20 \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & {{u}_{1}}q+{{u}_{1}}{{q}^{4}}-{{u}_{1}}{{q}^{3}}=10 \\ & {{u}_{1}}{{q}^{2}}+{{u}_{1}}{{q}^{5}}-{{u}_{1}}{{q}^{4}}=20 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & {{u}_{1}}q\left( 1+{{q}^{3}}-{{q}^{2}} \right)=10 \\ & {{u}_{1}}{{q}^{2}}\left( 1+{{q}^{3}}-{{q}^{2}} \right)=20 \\ \end{aligned} \right.\\&\Leftrightarrow \left\{ \begin{aligned} & q=2 \\ & {{u}_{1}}.2.(1+2^3-2^2)=10 \\ \end{aligned} \right.\\& \Leftrightarrow \left\{ \begin{aligned} & q=2 \\ & {{u}_{1}}=1 \\ \end{aligned} \right. \\ \end{aligned} \)
