Ta có: A= $x$+$\frac{9}{2x-1}$+$9$
⇔ 2A= $2x$+$\frac{18}{2x-1}$+$18$
= $2x-1$+$\frac{18}{2x-1}$+$17$
Áp dụng bđt cô si, ta có: A≥ 2$\sqrt[]{(2x-1).\frac{18}{2x-1}}$+$17$= 6.$\sqrt[]{2}$+$17$
Dấu = xảy ra khi $2x-1$=$\frac{18}{2x-1}$
⇔ $(2x-1)²$= $18$
⇔ $2x-1$= $±\sqrt[]{18}$
⇔ $x$= $\frac{\sqrt[]{18}+1}{2}$ ( vì x≥ $\frac{1}{2}$ )
Vậy Amin= 6.$\sqrt[]{2}$+$17$ khi $x$= $\frac{\sqrt[]{18}+1}{2}$