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Đặt A=3|x-5|-|3x+2|
TH1: \(x<-\frac23\)
=>3x+2<0; x-5<0
=>\(A=3\left(5-x\right)-\left(-3x-2\right)\)
=15-3x+3x+2
=17
TH2: \(-\frac23\le x<5\)
=>3x+2>=0; x-5<0
=>\(A=3\left(5-x\right)-\left(3x+2\right)\)
=15-3x-3x-2
=-6x+13
TH3: x>=5
=>3x+2>0; x-5>=0
=>A=3(x-5)-(3x+2)
=3x-15-3x-2
=-17


Ta có : \(\hept{\begin{cases}\left|x+3\right|\ge0\forall x\\\left|4-x\right|\ge0\forall x\\\left|x\right|\ge0\forall x\end{cases}\Rightarrow}\hept{\begin{cases}\left|x+3\right|=x+3\\\left|4-x\right|=4-x\\\left|x\right|=x\end{cases}}\)
\(\Rightarrow3\left|x-3\right|+2\left|4-x\right|+\left|x\right|\)
\(=3.\left(x-3\right)+2.\left(4-x\right)+x\)
\(=3x-9+8-2x+x\)
\(=\left(3x-2x+x\right)-\left(9-8\right)\)
\(=2x+1\)




+ Nếu x < -3, ta có:
3(x - 1) - 2|x + 3| = 3(x - 1) - 2(-3 - x)
= 3x - 3 + 6 + 2x
= 5x + 3
+ Nếu \(x\ge-3\) ta có:
3(x - 1) - 2|x + 3| = 3(x - 1) - 2(x + 3)
= 3x - 3 - 2x - 6
= x - 9
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