\(A=x-\sqrt{x}+1\)
\(A=x-2.\sqrt{x}.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(A=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\)
có \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\forall x\) => \(\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\) => \(A\ge\frac{3}{4}\forall x\)
vậy \(A_{min}=\frac{3}{4}< =>\sqrt{x}-\frac{1}{2}=0< =>\sqrt{x}=\frac{1}{2}< =>x=\frac{1}{4}\)