\(\Leftrightarrow\left\{{}\begin{matrix}u_1+u_1q^2+u_1q^4=-21\\u_1q+u_1q^3=10\end{matrix}\right.\)
Chia vế cho vế:
\(\frac{1+q^2+q^4}{q+q^3}=-\frac{21}{10}\)
\(\Leftrightarrow10q^4+21q^3+10q^2+21q+10=0\)
Nhận thấy \(q=0\) không phải là nghiệm, chia 2 vế cho \(q^2\):
\(10\left(q^2+\frac{1}{q^2}\right)+21\left(q+\frac{1}{q}\right)+10=0\)
Đặt \(q+\frac{1}{q}=x\) với \(\left|x\right|\ge2\Rightarrow q^2+\frac{1}{q^2}=x^2-2\)
\(\Rightarrow10\left(x^2-2\right)+21x+10=0\)
\(\Leftrightarrow10x^2+21x-10=0\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{2}{5}\left(l\right)\\x=-\frac{5}{2}\end{matrix}\right.\)
\(\Rightarrow q+\frac{1}{q}=-\frac{5}{2}\Leftrightarrow2q^2+5q+2=0\Rightarrow\left[{}\begin{matrix}q=-2\\q=-\frac{1}{2}\end{matrix}\right.\)
- Với \(q=-2\Rightarrow u_1=-1\)
- Với \(q=-\frac{1}{2}\Rightarrow u_1=-16\)
