\(|x+1|+|x+2|+|x+3|+|x+3|+|x+4|=x-5\left(1\right)\)
Ta có: \(|x+1|\ge0;\forall x\)
\(|x+2|\ge0;\forall x\)
\(|x+3|\ge0;\forall x\)
\(|x+4|\ge0;\forall x\)
\(\Rightarrow|x+1|+|x+2|+|x+3|+|x+3|+|x+4|\ge0;\forall x\)
Mà \(|x+1|+|x+2|+|x+3|+|x+3|+|x+4|=x-5\)
\(\Rightarrow x-5\ge0\)
\(\Rightarrow x\ge5\)
\(\Rightarrow|x+1|=x+1\)
\(|x+2|=x+2\)
..................................
\(|x+4|=x+4\)
Thay vào (1) ta được :
\(x+1+x+2+x+3+x+3+x+4=x-5\)
\(\Leftrightarrow5x+13=x-5\)
\(\Leftrightarrow4x=-18\)
\(\Leftrightarrow x=\frac{-9}{2}\)
Vậy \(x=\frac{-9}{2}\)
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