ĐKXĐ: x>=-1/3
Ta có: \(x^2+x-4\sqrt{3x+1}+6=0\)
=>\(x^2-x+2x-2-4\sqrt{3x+1}+8=0\)
=>\(\left(x-1\right)\left(x+2\right)-4\left(\sqrt{3x+1}-2\right)=0\)
=>\(\left(x-1\right)\left(x+2\right)-4\cdot\frac{3x+1-4}{\sqrt{3x+1}+2}=0\)
=>\(\left(x-1\right)\left(x+2\right)-4\cdot\frac{3x-3}{\sqrt{3x+1}+2}=0\)
=>\(\left(x-1\right)\left(x+2-\frac{12}{\sqrt{3x+1}+2}\right)=0\)
=>x-1=0
=>x=1(nhận)

