Theo bài ra ta có: \(\left|x-4\right|=\left|x+3,1\right|\)
=> x-4 = \(\pm\) (x+3,1)
=>\(\left[\begin{array}{nghiempt}x-4=x+3,1\\x-4=-x-3,1\end{array}\right.\) => \(\left[\begin{array}{nghiempt}x-x=3,1+4\\x+x=-3,1+4\end{array}\right.\)
=> \(\left[\begin{array}{nghiempt}0=7,1\left(loại\right)\\2x=0,9\end{array}\right.\) => \(\left[\begin{array}{nghiempt}x\in\varnothing\\x=0,45\end{array}\right.\)
Vậy x=0,45