𝓓𝓾𝔂 𝓐𝓷𝓱 tr ơi ghê chưa ghê chưa =.=

\(P=\dfrac{\sqrt{x}-2}{\sqrt{x}}=1-\dfrac{2}{\sqrt{x}}\)

Vì \(x\le3\Rightarrow\dfrac{2}{\sqrt{x}}\ge\dfrac{2}{\sqrt{3}}\)\(\Leftrightarrow-\dfrac{2}{\sqrt{x}}\le-\dfrac{2}{\sqrt{3}}\)\(\Leftrightarrow1-\dfrac{2}{\sqrt{3}}\le1-\dfrac{2}{\sqrt{3}}\)

\(\Rightarrow\)\(P\le\dfrac{3-2\sqrt{3}}{3}\)

Dấu = xra khi x=3

Vậy \(P_{max}=\dfrac{3-2\sqrt{3}}{3}\)

a)\(\overrightarrow{AB}+\overrightarrow{BO}+\overrightarrow{OA}=\overrightarrow{AA}=\overrightarrow{0}\)

b)\(\overrightarrow{BC}+\overrightarrow{OA}+\overrightarrow{OD}=\overrightarrow{OA}+\overrightarrow{AD}+\overrightarrow{OD}=\overrightarrow{OD}+\overrightarrow{OD}=2\overrightarrow{OD}\)

c)\(\overrightarrow{OA}+\overrightarrow{BC}+\overrightarrow{DO}+\overrightarrow{CD}=\left(\overrightarrow{DO}+\overrightarrow{OA}\right)+\left(\overrightarrow{BC}+\overrightarrow{CD}\right)=\overrightarrow{DA}+\overrightarrow{BD}=\overrightarrow{BA}\)