\(|x+1|+|x+2|+|x+3|=4x-4\left(1\right)\)
Vì \(\hept{\begin{cases}|x+1|\ge0;\forall x\\|x+2|\ge0;\forall x\\|x+3|\ge0;\forall x\end{cases}}\)
\(\Rightarrow|x+1|+|x+2|+|x+3|\ge0;\forall x\)
Mà \(|x+1|+|x+2|+|x+3|=4x-4\)
\(\Rightarrow4x-4\ge0\)
\(\Rightarrow4x\ge4\)
\(\Rightarrow x\ge1>0\)
\(\Rightarrow\hept{\begin{cases}|x+1|=x+1\\|x+2|=x+2\\|x+3|=x+3\end{cases}}\)(2)
Thay (2) vào (1) ta được :
\(x+1+x+2+x+3=4x-4\)
\(\Leftrightarrow3x+6=4x-4\)
\(\Leftrightarrow3x-4x=-4-6\)
\(\Leftrightarrow-x=-10\)
\(\Leftrightarrow x=10\)
Vậy x=10
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