Ta có: \(\left(3x-1\right)\left(x^2+4\right)-\left(3x-1\right)^2=0\)
=>\(\left(3x-1\right)\left(x^2+4-3x+1\right)=0\)
=>\(\left(3x-1\right)\left(x^2-3x+5\right)=0\)
mà \(x^2-3x+5=x^2-3x+\dfrac{9}{4}+\dfrac{11}{4}=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}>=\dfrac{11}{4}>0\)
nên 3x-1=0
=>3x=1
=>\(x=\dfrac{1}{3}\)