\(\left|x+3\right|\ge0;\left|x+1\right|\ge0\)
\(\Rightarrow\left|x+3\right|+\left|x+1\right|\ge0\)
Để \(\left|x+3\right|+\left|x+1\right|=3x\) thì \(3x\ge0\Rightarrow x\ge0\)
\(\Rightarrow x+3>0\) và \(x+5>0\)
Ta có :
\(x+3+x+1=3x\)
\(\Rightarrow2x+\left(3+1\right)=3x\)
\(2x+4=3x\)
\(\Leftrightarrow4=3x-2x\)
\(\Rightarrow4=1x\)
\(\Leftrightarrow x=4\)
Vậy \(x=4\) .
Ta thấy:\(VT\ge0\Rightarrow VP\ge0\Rightarrow3x\ge0\Rightarrow x\ge0\)
Khi đó \(\left|x+3\right|+\left|x+1\right|=3x\)
\(\Leftrightarrow\left(x+3\right)+\left(x+1\right)=3x\)
\(\Rightarrow2x+4=3x\Rightarrow x=4\) (thỏa mãn)
