\(x^2-x-1=0\) có \(\Delta=\left(-1\right)^2-4.1.\left(-1\right)=5\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=\frac{1+\sqrt{5}}{2}\\x_2=\frac{1-\sqrt{5}}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x_1=\frac{1-\sqrt{5}}{2}\\x_2=\frac{1+\sqrt{5}}{2}\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}S=\left(\frac{1+\sqrt{5}}{2}\right)^8+\left(\frac{1-\sqrt{5}}{2}\right)^6+13\left(\frac{1-\sqrt{5}}{2}\right)=39\\S=\left(\frac{1-\sqrt{5}}{2}\right)^8+\left(\frac{1+\sqrt{5}}{2}\right)^6+13\left(\frac{1+\sqrt{5}}{2}\right)=39\end{matrix}\right.\)
Vậy \(S=39\)
Theo hệ thức Vi-et ta có \(x_1+x_2=1\)
Vì x1,x2 là nghiệm của PT nên
\(x_1^2=x_1+1\)\(\Leftrightarrow x^4_1=x_1^2+2x_1+1=3x_1+2\)
\(\Leftrightarrow x^8_1=9x_1^2+12x_1+4=9\left(x_1+1\right)+12x_1+4\)\(=21x_1+13\)
\(x_2^6=\left(x_2+1\right)\left(3x_2+2\right)=3x_2^2+5x_2+2=8x_2+5\)
\(\Rightarrow S=21x_1+13+8x_2+5+13x_2\)
\(=21\left(x_1+x_2\right)+18=21+18=39\)
