|x+12|+|x+13|+|x+14|=4x
|x+12|+|x+13|+|x+14| luôn\(\ge\) 0=>x \(\ge\)0
|x+12|+|x+13|+|x+14| là số dương hay|x+12|+|x+13|+|x+14| =x+12+x+13+x+14=4x
x+12+x+13+x+14=4x
3x+(12+13+14)=4x
4x-3x=12+13+14
x=39
Do \(\left|x+12\right|\ge0,\left|x+13\right|\ge0,\left|x+14\right|\ge0\)
\(\Rightarrow\left|x+12\right|+\left|x+13\right|+\left|x+14\right|>0\)
Do VP>0 suy ra VT>0
Hay \(4x>0\)
\(\Rightarrow x>0\)
Khi đó:\(\left|x+12\right|+\left|x+13\right|+\left|x+14\right|=4x\)
\(\Leftrightarrow3x+39=4x\)(vì x>0)
\(\Rightarrow x=39\)
\(\left|x+12\right|+\left|x+13\right|+\left|x+14\right|=4x^{\left(1\right)}\)
Vì \(\left|x+12\right|\ge0;\left|x+13\right|\ge0;\left|x+14\right|\ge0\)
\(\Rightarrow\left|x+12\right|+\left|x+13\right|+\left|x+14\right|\ge0\)
Mà \(VT\ge0\Rightarrow VP\ge0\Rightarrow4x\ge0\Rightarrow x\ge0\)
Từ \(\left(1\right)\Rightarrow\left|x+12\right|+\left|x+13\right|+\left|x+14\right|=4x\)
\(\Rightarrow3.x+39=4x\)
\(\Rightarrow39=4x-3x\)
\(\Rightarrow x=39\)
