Công bội nguyên dương của cấp số nhân

B

- Hướng dẫn giải

Chọn B

Ta giải hệ: 

$\left\{\begin{array}{l}{u}_1\left(q^2+q+1\right)=21\\ \frac{1}{u_1}\left(1+\frac{1}{q}+\frac{1}{q^2}\right)=\frac{7}{12}\end{array}\right.$

$\begin{array}{l}\iff \left\{\begin{array}{l}\frac{q^2+q+1}{21}=\frac{1}{u_1}\\ \frac{1}{u_1}\left(\frac{q^2+q+1}{q^2}\right)=\frac{7}{12}\end{array}\right.\\ \iff \left\{\begin{array}{l}\frac{q^2+q+1}{21}=\frac{1}{u_1}\\ \frac{q^2+q+1}{21}.\frac{q^2+q+1}{q^2}=\frac{7}{12}\left(*\right)\end{array}\right.\end{array}$

Giải (*)

$\begin{array}{l}\frac{{(q^2+q+1)}^2}{q^2}=\frac{49}{4}\\ \iff \frac{q^2+q+1}{q}=\pm \frac{7}{2}\end{array}$

TH1:

$$\begin{gathered} \frac{q^2+q+1}{q}=\frac{7}{2}\iff 2\hspace{0.17em}{q}^2+\hspace{0.17em}2q+\hspace{0.17em}2 = 7q \\ \iff 2q^2- 5q +\hspace{0.17em} 2= 0 \\ \iff \left[\begin{array}{l}q=2\\ q=\frac{1}{2}\left(loai\right)\end{array}\right. \end{gathered}$$

TH2;

$$\begin{gathered} \frac{q^2+q+1}{q}=-\frac{7}{2}\iff 2q^2+\hspace{0.17em}2q + 2 = -7q \\ \iff 2q^2+\hspace{0.17em}9q+\hspace{0.17em}2 = 0 \\ \iff \left[\begin{array}{l}q=\frac{-9+\sqrt{65}}{4}\\ q=\frac{-9-\sqrt{65}}{4}\end{array}\right.\left(loai\right) \end{gathered}$$

Vậy q=2