Do \(x=0\) không phải nghiệm
\(x^2+3x+1=0\Leftrightarrow x+3+\frac{1}{x}=0\Leftrightarrow x+\frac{1}{x}=-3\)
\(\Leftrightarrow\left(x+\frac{1}{x}\right)^2=9\Rightarrow x^2+\frac{1}{x^2}=7\)
Đặt \(x_n=x^n+\frac{1}{x^n}\Rightarrow x_1=-3;x_2=7\)
\(x_1x_n=\left(x+\frac{1}{x}\right)\left(x^n+\frac{1}{x^n}\right)=x^{n+1}+\frac{1}{x^{n+1}}+x^{n-1}+\frac{1}{x^{n-1}}=x_{n+1}+x_{n-1}\)
\(\Rightarrow x_{n+1}=x_1x_n-x_{n-1}=-3x_n-x_{n-1}\)
Cho \(n=2\Rightarrow x_3=x^3+\frac{1}{x^3}=-3.x_2-x_1=-18\)
\(n=3\Rightarrow x_4=x^4+\frac{1}{x^4}=-3x_3-x_2=47\)
\(n=4\Rightarrow x_5=x^5+\frac{1}{x^5}=-3x_4-x_3=-123\)
\(n=5\Rightarrow x_6=x^6+\frac{1}{x^6}=-3x_5-x_4=322\)
Thay vào và tính, kết quả rất to
