\(Q=x-2\sqrt{2x-1}=\dfrac{2x-4\sqrt{2x-1}}{2}\) với \(x\ge\dfrac{1}{2}\)
\(=\dfrac{2x-1-4\sqrt{2x-1}+4-3}{2}\)
\(\dfrac{\left(\sqrt{2x-1}-2\right)^2-3}{2}\)
Ta có : \(\left(\sqrt{2x-1}-2\right)^2\ge0\) với \(\forall x\ge\dfrac{1}{2}\)
=> \(\left(\sqrt{2x-1}-2\right)^2-3\ge-3\)
=> \(Q\ge\dfrac{-3}{2}\) Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{2x-1}-2\right)^2=0\Leftrightarrow2x-1=4\Leftrightarrow x=2,5\)( Thỏa mãn ĐK)
Vậy MinQ=\(\dfrac{-3}{2}\) \(\Leftrightarrow x=2,5\)