Vì \(\left|x+1\right|\ge0,\left|x+2\right|\ge0,\left|x+3\right|\ge0\)
\(\Rightarrow4x\ge0\)
Phương trình đã cho trở thành:
\(x+1+x+2+x+3=4x\)
\(\Rightarrow3x-4x=-3-2-1\)
\(\Rightarrow-x=-6\)
\(\Rightarrow x=6\)
ta có | x+1| +|x+2|+|x+3|=4x
do |x+1| ;|x+2|;|x+3| \(\ge0\) mà |x+1|+|x+2|+|x+3|=4x
=>4x\(\ge0\) \(\Rightarrow x\ge0\)
\(\Rightarrow x+1+x+2+x+3=4x\)
\(\Rightarrow3x+6=4x\) \(\Rightarrow4x-3x=6\Rightarrow x=6\)
Ta có: \(|x+1|+|x+2|+\left|x+3\right|=4x\)
\(\Rightarrow x+1+x+2+x+3=4x\)
\(3x+6=4x\)
Có hai trường hợp:
\(3x+6=4x\) \(3x+6=-\left(4x\right)\)
\(\Rightarrow3x-4x=-6\) \(\Rightarrow3x+4x=-6\)
\(-x=-6\) \(7x=-6\)
\(x=6\) \(x=-\dfrac{6}{7}\)
Vậy: \(x=6\) hoặc \(x=-\dfrac{6}{7}\)
