\(2\left|x-1\right|+3\ge3\)
\(\Leftrightarrow\dfrac{1}{2\left|x-1\right|+3}\le\dfrac{1}{3}\)
Dấu "=" xảy ra khi x=1
\(D=\dfrac{1}{2\left|x-1\right|+3}\)
Ta có:\(\left|x-1\right|\ge0\forall x\)
\(\Rightarrow2\left|x-1\right|\ge0\forall x\)
\(\Rightarrow2\left|x-1\right|+3\ge3\forall x\)
\(\Rightarrow\dfrac{1}{2\left|x-1\right|+3}\le\dfrac{1}{3}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x-1=0\)
\(\Rightarrow x=1\)
Vậy \(max\) \(E=\dfrac{1}{3}\) khi \(x=1\)