\(y=\dfrac{6\sqrt{2}.x.\dfrac{1}{6\sqrt{2}}+\dfrac{2\sqrt{2}}{3}.\dfrac{3\sqrt{2}}{4}\sqrt{1+9x^2}}{8x^2+1}\)

\(y\le\dfrac{3\sqrt{2}\left(\dfrac{1}{72}+x^2\right)+\dfrac{\sqrt{2}}{3}\left(\dfrac{9}{8}+9x^2+1\right)}{8x^2+1}=\dfrac{\sqrt{2}\left(6x^2+\dfrac{3}{4}\right)}{8x^2+1}\)

\(y\le\dfrac{\dfrac{3\sqrt{2}}{4}\left(8x^2+1\right)}{8x^2+1}=\dfrac{3\sqrt{2}}{4}\)

ta có :  (\(\sqrt{x}\)-   2   )\(^2\)\(\ge\)0

\(\Leftrightarrow\)x  -  4\(\sqrt{x}\)+  4  \(\ge\)0

\(\Leftrightarrow\)x  -  4\(\sqrt{x}\)+  4 +   8\(\sqrt{x}\) \(\ge\)8\(\sqrt{x}\)

   \(\Leftrightarrow\)(\(\sqrt{x}\)+    2  )\(^2\)\(\ge\)8\(\sqrt{x}\)

\(\Leftrightarrow\)-(\(\sqrt{x}\)+    2  )\(^2\)\(\le\)-8\(\sqrt{x}\)

\(\Leftrightarrow\)Q  \(\le\)\(\frac{-8\sqrt{x}}{\sqrt{x}}\)=   (   -  8  )

Dấu ''   =   ''   xaye ra tại   x =  4

ĐKXĐ: ...

\(A=-\left(2-x\right)+\sqrt{2-x}-\frac{1}{4}+\frac{9}{4}\)

\(A=-\left(2-x-\sqrt{2-x}+\frac{1}{4}\right)+\frac{9}{4}\)

\(A=-\left(\sqrt{2-x}-\frac{1}{2}\right)^2+\frac{9}{4}\le\frac{9}{4}\)

\(A_{max}=\frac{9}{4}\) khi \(\sqrt{2-x}=\frac{1}{2}\Rightarrow x=\frac{7}{4}\)

bạn đặt ĐKXĐ và rút gọn P đi\(\sqrt{x}-x=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4},\forall x\ne1\)

\(\Rightarrow Maxp=\frac{1}{4}\Leftrightarrow dấu=xảyra\)

\(\Leftrightarrow x=\frac{1}{4}\)

Câu 2: 

\(C=-x+\sqrt{x}\)

\(=-\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{1}{4}\)

\(=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{4}\)