TH1 : \(x< -4;\)ta có:
\(-\left(x+1\right)+\left[-\left(x+4\right)\right]=3x\)
\(-2x-5=3x\)
\(-2x-3x=5\)
\(-5x=5\)
\(x=-1\)(không thỏa mãn )
TH2 : \(-4\le x< -1;\)ta có:
\(-\left(x+1\right)+\left(x+4\right)=3x\)
\(3=3x\)
\(x=1\)( không thỏa mãn)
TH3 : \(x\ge-1;\)ta có:
\(\left(x+1\right)+\left(x+4\right)=3x\)
\(2x+5=3x\)
\(x=5\)(thỏa mãn)
Vậy \(x=5\).