Điều kiện áp dụng BĐT Cauchy:$x \ge 0$
$x+\dfrac{2}{2x-1}\\ =\dfrac{1}{2}(2x-1)+\dfrac{2}{2x-1}+\dfrac{1}{2}\\ \ge 2\sqrt{\dfrac{1}{2}(2x-1).\dfrac{2}{2x-1}}+\dfrac{1}{2}\left(x \ge 0,x\ne\dfrac{1}{2}\right)\\ =2+\dfrac{1}{2}\\ =\dfrac{5}{2}$
Dấu "=" xảy ra $\Leftrightarrow \dfrac{1}{2}(2x-1)=\dfrac{2}{2x-1}\Leftrightarrow x=\dfrac{3}{2};x=-\dfrac{1}{2}(L)$
