Ta có: \(x^2-x-1=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}-\frac{5}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2-\frac{5}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=\frac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{2}=\sqrt{\frac{5}{4}}\\x-\frac{1}{2}=-\sqrt{\frac{5}{4}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1+\sqrt{5}}{2}\\x=\frac{1-\sqrt{5}}{2}\end{matrix}\right.\)
Vậy: \(x=\frac{1\pm\sqrt{5}}{2}\)
x2 - x - 1 = 0
\(\Leftrightarrow\) (x - \(\frac{1}{2}\))2 - \(\frac{5}{4}\) = 0
\(\Leftrightarrow\) (x - \(\frac{1}{2}\) - \(\sqrt{\frac{5}{4}}\))(x - \(\frac{1}{2}\) + \(\sqrt{\frac{5}{4}}\)) = 0
\(\Leftrightarrow\) (x - \(\frac{1-\sqrt{5}}{2}\))(x - \(\frac{1+\sqrt{5}}{2}\)) = 0
\(\Leftrightarrow\) x - \(\frac{1-\sqrt{5}}{2}\) = 0 hoặc x - \(\frac{1+\sqrt{5}}{2}\) = 0
\(\Leftrightarrow\) x = \(\frac{1-\sqrt{5}}{2}\) và x = \(\frac{1+\sqrt{5}}{2}\)
Vậy S = {\(\frac{1-\sqrt{5}}{2}\); \(\frac{1+\sqrt{5}}{2}\)}
Chúc bn học tốt!!
