$a)$ 

$ 3x^2 -x -1 = 0 \to 3(x^2 - \dfrac{1}{3}x - \dfrac{1}{3}) = 0$

$ \to x^2 - \dfrac{1}{3}x - \dfrac{1}{3} = 0$

$ \to x^2 -\dfrac{1- \sqrt{13}}{6}x - \dfrac{1+ \sqrt{13}}{6}x - \dfrac{1- \sqrt{13}}{6}. \dfrac{1+ \sqrt{13}}{6}$

$\to x(x - \dfrac{1- \sqrt{13}}{6}) - \dfrac{1+ \sqrt{13}}{6}( x - \dfrac{1- \sqrt{13}}{6})$

$ \to (x - \dfrac{1- \sqrt{13}}{6}) ( x - \dfrac{1+ \sqrt{13}}{6}) = 0$

$ \to$ \(\left[ \begin{array}{l}x= \dfrac{1- \sqrt{13}}{6}\\x=\dfrac{1+ \sqrt{13}}{6}\end{array} \right.\) 

$b)$

$ x^2 + 2x  -2 = 0 $

$\to x^2 + (\sqrt{3}+1)x - (\sqrt{3}-1)x - (\sqrt{3}+1)(\sqrt{3}-1) =0$

$ \to x(x + \sqrt{3}+1) - (\sqrt{3}-1)(x+ \sqrt{3}+1) = 0$

$ \to [x - (\sqrt{3}-1)](x + \sqrt{3}+1) = 0$

$ \to$ \(\left[ \begin{array}{l}x= \sqrt{3}-1\\x=-1 -\sqrt{3} \end{array} \right.\) 

$c)$

$ x^2- 4x -6 = 0$

$\to x^2 - (2-\sqrt{10})x - (2+ \sqrt{10}x + (2-\sqrt{10})(2+ \sqrt{10}) = 0$

$\to x[x - (2-\sqrt{10})] - (2+ \sqrt{10})[x - (2-\sqrt{10})] = 0$

$\to[x- (2+ \sqrt{10})][x - (2-\sqrt{10})] = 0$

$ \to$ \(\left[ \begin{array}{l}x=2+ \sqrt{10} \\x=2-\sqrt{10}\end{array} \right.\) 

Đáp án+Giải thích các bước giải:

`a)3x^2-x-1=0`

`<=>3(x^2-1/3x-1/3)=0`

`<=>x^2-1/3x-1/3=0`

`<=>x^2-2.x.\frac{1}{6}+(1/6)^2-13/36`=0`

`<=>(x-1/6)^2-(\frac{\sqrt{13}}{6})^2`

`<=>(x-1/6-\frac{\sqrt{13}}{6})(x-1/6+\frac{\sqrt{13}}{6})=0`

`TH1:`

`x-1/6-\frac{\sqrt{13}}{6}=0`

`<=>x-1/6=0+\frac{\sqrt{13}}{6}`

`<=>x=\frac{\sqrt{13}}{6}+1/6`

`<=>x=\frac{\sqrt{13}+1}{6}`

`TH2:`

`x-1/6+\frac{\sqrt{13}}{6}=0`

`<=>x-1/6=0-\frac{\sqrt{13}}{6}`

`<=>x=-\frac{\sqrt{13}}{6}+1/6`

`<=>x=\frac{1-\sqrt{13}}{6}`

 Vậy `S=\{\frac{\sqrt{13}+1}{6};\frac{1-\sqrt{13}}{6}\}`

`b)x^2+2x-2=0`

`<=>x^2+2x+1-3=0`

`<=>x^2+2.x.1+1^2-3=0`

`<=>(x-1)^2-(\sqrt{3})^2=0`

`<=>(x-1-\sqrt{3})(x-1+\sqrt{3})=0`

`TH1:`

`x-1-\sqrt{3}=0`

`<=>x-1=0+\sqrt{3}`

`<=>x=\sqrt{3}+1`

`TH2:`

`x-1+\sqrt{3}=0`

`<=>x-1=0-\sqrt{3}`

`<=>x=-\sqrt{3}+1`

`<=>x=1-\sqrt{3}`

 Vậy `S=\{\sqrt{3}+1;1-\sqrt{3}\}`

`c)x^2-4x-6=0`

`<=>x^2-4x+4-10=0`

`<=>(x-2)^2-(\sqrt{10})^2=0`

`<=>(x-2-\sqrt{10})(x-2+\sqrt{10})=0`

`TH1:x-2-\sqrt{10}=0`

`<=>x-2=0+\sqrt{10}`

`<=>x=\sqrt{10}+2`

`TH2:x-2+\sqrt{10}=0`

`<=>x-2=0-\sqrt{10}`

`<=>x=-\sqrt{10}+2`

`<=>x=2-\sqrt{10}`

 Vậy `S=\{\sqrt{10}+2;2-\sqrt{10}\}`