a) A = x - $\sqrt[2]{x}$ + 3 (x ≥ 0)
= x - $\sqrt[2]{x}$ + $\frac{1}{4}$ + $\frac{11}{4}$
= ($\sqrt[2]{x}$ - $\frac{1}{2}$)² + $\frac{11}{4}$
Có: ($\sqrt[2]{x}$ - $\frac{1}{2}$)² ≥ 0 với mọi x ≥ 0
⇒ ($\sqrt[2]{x}$ - $\frac{1}{2}$)² + $\frac{11}{4}$ ≥ $\frac{11}{4}$
⇒ A ≥ $\frac{11}{4}$
Dấu"=" xảy ra ⇔ $\sqrt[2]{x}$ - $\frac{1}{2}$ = 0
⇔ $\sqrt[2]{x}$ = $\frac{1}{2}$
⇔ x = $\frac{1}{4}$ (t/m)
Vậy min A = $\frac{11}{4}$ khi x = $\frac{1}{4}$
