Có: \(\begin{cases}\left|x+1\right|\ge0\\\left|x+2\right|\ge0\\\left|x+3\right|\ge0\\\left|x+4\right|\ge0\end{cases}\)\(\forall x\)
Do đó, \(5x-1\ge0\Rightarrow5x\ge1\Rightarrow x\ge\frac{1}{5}\)
Lúc này ta có: \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)=5x-1\)
=> 4x + 10 = 5x - 1
=> 10 + 1 = 5x - 4x
=> x = 11
Vậy x = 11