\(\left|x+1\right|+\left|x+4\right|=3x\)
TH1 : \(x< -4\) ; ta có :
\(-\left(x+1\right)+\left[-\left(x+4\right)\right]=3x\)
\(-2x+5=3x\)
\(-2x-3x=5\)
\(-5x=5\)
\(x=-1\left(KTM\right)\)
TH2 : \(-4\le x< -1\) ; ta có :
\(-\left(x+1\right)+\left(x+4\right)=3x\)
\(3=3x\)
\(x=1\left(KTM\right)\)
TH3 : \(x\ge-1\) ; ta có :
\(\left(x+1\right)\left(x+4\right)=3x\)
\(2x+5=3x\)
\(x=5\left(TM\right)\)
Vậy \(x=5\)
Ta có : \(\left|x+1\right|+\left|x+4\right|=3x\)
Vì vế trái luôn dương nên x > 0
Suy ra pt trở thành \(\left(x+1\right)+\left(x+4\right)=3x\Leftrightarrow x=5\)