1) x2 - 6x + 7 = 0
⇔ x2 - 2.3.x + 9 - 2 = 0
⇔ (x - 3)2 = 2
⇔ \(\left\{{}\begin{matrix}x-3=\sqrt{2}\\x-3=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{2}+3\\x=-\sqrt{2}+3\end{matrix}\right.\)
Vậy x = \(\sqrt{2}+3\) và x = \(-\sqrt{2}+3\)
3) 9x2 + x + 1 = 0
⇔ (3x)2 + 2.3x.\(\dfrac{1}{6}\) + \(\left(\dfrac{1}{6}\right)^2+\dfrac{35}{36}\) = 0
⇔ \(\left(3x+\dfrac{1}{6}\right)^2\) = \(-\dfrac{35}{36}\)
Vì \(\left(3x+\dfrac{1}{6}\right)^2\ge0\) nên suy ra x không có.
2) x2 - x + 1= 0
⇔ x2 - 2.\(\dfrac{1}{2}\).x + \(\dfrac{1}{4}+\dfrac{3}{4}\)= 0
⇔ \(\left(x-\dfrac{1}{2}\right)^2\)= \(-\dfrac{3}{4}\)
Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\) nên suy ra x không có.
