[tex]\sqrt{3x+1}-\sqrt{6-x}+3x^2-14x-8=0\Rightarrow \sqrt{3x+1}-4+(1-\sqrt{6-x})+3x^2-14x-5=0\Leftrightarrow \frac{3x-15}{\sqrt{3x+1}+4}+\frac{x-5}{\sqrt{6-x}+1}+(x-5)(3x+1)=0\Leftrightarrow (x-5)(\frac{3}{\sqrt{3x+1}+4}+\frac{1}{\sqrt{6-x}+1}+3x+1)=0[/tex]
Vì [TEX]\frac{3}{\sqrt{3x+1}+4}+\frac{1}{\sqrt{6-x}+1}+3x+1>0[/TEX] nên x = 5.