$\sqrt[]{9x^2+6x+1}$ = $\sqrt[]{11-6\sqrt[]{2}}$ giải phương trình trên

`\sqrt{9x^2+6x+1}=\sqrt{11-6\sqrt{2}}` ĐK: `9x^2+6x+1>=0` với `AAx`

`<=> \sqrt{(3x+1)^2}=\sqrt{2-2.3\sqrt{2}+9}`

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

`<=> |3x+1|=|\sqrt{2}-3|`

`<=> |3x+1|=3-\sqrt{2}`

`<=>`\(\left[ \begin{array}{l}3x+1=3-\sqrt{2}\\3x+1=\sqrt{2}-3\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}3x=2-\sqrt{2}\\3x=\sqrt{2}-4\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=\dfrac{2-\sqrt{2}}{3}\\x=\dfrac{\sqrt{2}-4}{3}\end{array} \right.\) 

Vậy `S={ (2-\sqrt{2})/3; (\sqrt{2}-4)/3 }`