\(x^3-3x^2+\sqrt{5}x=0\Leftrightarrow x\left(x^2-3x^2+\sqrt{5}\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x^2-3x+\sqrt{5}=0\left(1\right)\end{matrix}\right.\)
Từ phương trình (1) ta có △=\(b^2-4ac=\left(-3\right)^2-4.1.\sqrt{5}=9-4\sqrt{5}\)>0
Vậy phương trình (1) có 2 nghiệm phân biệt \(\left\{{}\begin{matrix}x_1=\frac{-b+\sqrt{\text{△}}}{2a}=\frac{3+\sqrt{9-4\sqrt{5}}}{2.1}=\frac{3+\sqrt{\left(\sqrt{5}-2\right)^2}}{2}=\frac{3+\sqrt{5}-2}{2}=\frac{\sqrt{5}+1}{2}\\x_2=\frac{-b-\sqrt{\text{△}}}{2a}=\frac{3-\sqrt{9-4\sqrt{5}}}{2.1}=\frac{3-\sqrt{\left(\sqrt{5}-2\right)^2}}{2}=\frac{3-\sqrt{5}+2}{2}=\frac{5-\sqrt{5}}{2}\end{matrix}\right.\)
Vậy S={0;\(\frac{\sqrt{5}+1}{2};\frac{5-\sqrt{5}}{2}\)}
