Ta có : \(\left|2x+1\right|+\left|x+3\right|\ge0\forall x\)
\(\Rightarrow5x\ge0\forall x\) ( Do \(\left|2x+1\right|+\left|x+3\right|=5x\) )
\(\Rightarrow x\ge0\)
\(\Rightarrow VT\) \(=\left|2x+1\right|+\left|x+3\right|\) sẽ trở thành : \(\left(2x+1\right)+\left(x+3\right)\)
\(\Rightarrow2x+1+x+3=5x\)
\(\Leftrightarrow3x+4=5x\)
\(\Leftrightarrow4=5x-3x\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)
Vậy : \(x=2\)
Có: \(\left|2x+1\right|\ge0,\forall x\)
\(\left|x+3\right|\ge0,\forall x\)
\(\Rightarrow5x\ge0\Rightarrow x\ge0\)
TH1:\(2x+1+x+3=5x\)
\(\rightarrow3x+4=5x\)
\(\rightarrow3x-5x=-4\)
\(\Rightarrow x=2\)
TH2:\(2x+1-x-3=5x\)
\(\rightarrow x-2=5x\)
\(\rightarrow x-5x=2\)
\(\Rightarrow x=\frac{-1}{2}\)(Loại, vì\(x\ge0\))
TH3:\(-2x-1+x+3=5x\)
\(\Rightarrow x=\frac{1}{3}\)
TH4:\(-2x-1-x-3=5x\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy\(x\in\left\{2;\frac{1}{3};\frac{1}{2}\right\}\)
