e)
Gọi công bội là q
Ta có: $\left \{ {{u_1+qu_1+q^2u_1=21(1)} \atop {\frac{1}{u_1}+\frac{1}{qu_1}+\frac{1}{q^2u_1}=\frac{7}{12}(2)}} \right.$
⇔$\left \{ {{u_1(1+q+q^2)=21} \atop {\frac{1}{u_1}(1+\frac{1}{q}+\frac{1}{q^2})=\frac{7}{12}}} \right.$
⇔$\left \{ {{u_1(1+q+q^2)=21} \atop {\frac{1}{u_1}(\frac{q^2+q+1}{q^2}=\frac{7}{12}}} \right.$
⇔$\left \{ {{u_1(1+q+q^2)=21} \atop {u_1^2q^2=36}} \right.$
TH1: $u_1q=-6⇔u_1=\frac{-6}{q}$ thế vào (1) ta được:
$-6q^2-27q-6=0$⇔...
TH2: $u_1q=-6⇔u_1=\frac{-6}{q}$ thế vào (1) ta được:
$6q^2-15q+6=0$⇔\(\left[ \begin{array}{l}q=2⇒u_1=3\\q=1/2⇒u_1=12\end{array} \right.\)
f) $\left \{ {{u_1(1+q+q^2+q^3)=30(1)} \atop {u_1^2(1+q^2+q^4+q^6)=340(2)}} \right.$
⇔$\left \{ {{u_1(1+q^2)(1+q)=30(1)} \atop {u_1^2(1+q^2)(1+q^4)=340(2)}} \right.$
Lấy $(1)^2/(2)$ ta được: $\frac{(1+q^2)(1+q)^2}{1+q^4}= \frac{45}{17}$
⇔$\frac{(1+q^2)(1+q^2+2q)}{(1+q^2)^2-2q^2}=\frac{45}{17}$
Đặt $1+q^2=t$ (t≥1)ta có:
$\frac{t(t+2q)}{(t^2-2q^2}=\frac{45}{17}$
⇔$17t^2+34qt=45t^2-90q^2$
⇔$28t^2-34qt-90q^2=0$
⇔\(\left[ \begin{array}{l}t=\frac{5}{2}q\\t=\frac{-9}{7}q\end{array} \right.\)
⇔\(\left[ \begin{array}{l}2+2q^2=5q\\7+7q^2=-9q\end{array} \right.\)
⇔\(\left[ \begin{array}{l}q=2⇒u_1=2\\q=1/2⇒u_1=16\end{array} \right.\)
