Lời giải:
a.
Ta có:
$-x^2-\sqrt{5}x=1$
$⇔-x^2-\sqrt{5}x-1=0$
Ta có:
$∆=b^2-4ac=(-\sqrt{5})^2-4.(-1).(-1)=1>0$=>Phương trình có 2 nghiệm phân biệt.
\(\left[ \begin{array}{l}x_1=\frac{-b+\sqrt{∆}}{2a}\\x_2=\frac{-b-\sqrt{∆}}{2a}\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x_1=\frac{-(-\sqrt{5})+\sqrt{1}}{2.(-1)}\\x_2=\frac{-(-\sqrt{5})-\sqrt{1}}{2.(-1)}\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x_1=\frac{1+\sqrt{5}}{-2}\\x_2=\frac{-1+\sqrt{5}}{-2}\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x_1=\frac{-1-\sqrt{5}}{2}\\x_2=\frac{1-\sqrt{5}}{2}\end{array} \right.\)
Vậy $S=${$\frac{-1-\sqrt{5}}{2};\frac{1-\sqrt{5}}{2}$}
