$P=\sqrt{x} -x\\P=\sqrt{x}-(\sqrt{x})²-\dfrac{1}{4}+\dfrac{1}{4}\\P=-[(\sqrt{x})²-\sqrt{x}+\dfrac{1}{4}]+\dfrac{1}{4}\\P=-[ (\sqrt{x})^2 - 2.\sqrt{x}.\dfrac{1}{2} + (\dfrac{1}{2})^2 ]+\dfrac{1}{4}\\P=-(\sqrt{x}-\dfrac{1}{2})^2+\dfrac{1}{4}$
Vì $-(\sqrt{x}-\dfrac{1}{2})^2≤0\\ ⇔ -(\sqrt{x}-\dfrac{1}{2})^2+\dfrac{1}{4}≤\dfrac{1}{4}$
Dấu $"="$ xảy ra khi $⇔\sqrt{x}-\dfrac{1}{2}=0\\⇔\sqrt{x}=\dfrac{1}{2}\\⇔x=\dfrac{1}{4}$
Vậy $P_{max}=\dfrac{1}{4}$ khi $x=\dfrac{1}{4}$
