`ĐKXĐ: 3x^2-8x-1>=0=>x>=(4+sqrt{19})/3`

`3x^2-6x+7=6 sqrt{2} \root{3}{x+1} \root{6}{3x^2-8x-1}`

`=6 sqrt{2} \root{6}{(x+1)^2(3x^2-8x-1)}`

`<= 6 sqrt{2} sqrt{3x^2-8x-1+x+1+x+1}*1/(sqrt3)`

`<= 2 sqrt{6} sqrt{3x^2-6x+1}`

Đặt `sqrt{3x^2-6x+1}=a`

`=> a^2+6<= 2 sqrt{6}a`

`=> a^2+6-2 sqrt{6}a<=0`

`=> a<= sqrt 6` 

`=>sqrt{3x^2-6x+1}<= sqrt 6`

`=>3x^2-6x+1<=6`

`=>3x^2-6x-5<=0`

`=>x<=(3+2sqrt6)/3`

Mà theo đkxđ `x>=(4+sqrt{19})/3`

`=> x \in \emptyset`