Bạn tham khảo nhé.
`3x^2-2\sqrt{6}x+1=0`
`<=>3x^2-2.\sqrt{3}x.\sqrt{2}+2-1=0`
`<=>(\sqrt{3}x)^2-2.\sqrt{3}x.\sqrt{2}+(\sqrt{2})^2-1=0`
`<=>(\sqrt{3}x-\sqrt{2})^2-1^2=0`
`<=>(\sqrt{3}x-\sqrt{2}-1)(\sqrt{3}x-\sqrt{2}+1)=0`
`<=>[(\sqrt{3}x-\sqrt{2}-1=0),(\sqrt{3}x-\sqrt{2}+1=0):}`
`<=>[(\sqrt{3}x=\sqrt{2}+1),(\sqrt{3}x=\sqrt{2}-1):}`
`<=>`$\left[\begin{matrix} x=\dfrac{\sqrt{2}+1}{\sqrt{3}} \\ x=\dfrac{\sqrt{2}-1}{\sqrt{3}}\end{matrix}\right.$
Vậy `x\in{\frac{\sqrt{2}+1}{\sqrt{3}};\frac{\sqrt{2}-1}{\sqrt{3}}}`
