ĐKXĐ : \(x\ge0\)

\(A=\frac{1}{5x+3\sqrt{x}+8}\le\frac{1}{5.0+3\sqrt{0}+8}=\frac{1}{8}\)

Dấu "=" xảy ra <=> x = 0

Vậy ...

ĐK: x > 0 

Vì x > 0

nên \(5x+3\sqrt{x}+8\ge0+0+8=0\)

\(\Rightarrow\frac{1}{5x+3\sqrt{x}+8}\le\frac{1}{8}\)

Dấu "='' <=> x = 0

ĐKXĐ: ...

\(A=\frac{1}{5\left(\sqrt{x}-\frac{3}{10}\right)^2+\frac{151}{10}}\le\frac{1}{\frac{151}{20}}=\frac{20}{151}\)

\(A_{max}=\frac{20}{151}\) khi \(\sqrt{x}=\frac{3}{10}\Rightarrow x=\frac{9}{100}\)

\(A=\frac{1}{5x-3\sqrt{x}+8}=\frac{1}{5\left(\sqrt{x}-\frac{3}{10}\right)^2+\frac{151}{20}}\le\frac{1}{\frac{151}{20}}=\frac{20}{151}\)

\(\Rightarrow A_{max}=\frac{20}{151}\) khi \(\sqrt{x}=\frac{3}{10}\Rightarrow x=\frac{9}{100}\)

ĐKXĐ :\(x\ge0\)

Mẫu :\(5x-3\sqrt{x}+8\)

\(=\left(\sqrt{5x}\right)^2-2.\frac{3\sqrt{5}}{10}.\sqrt{5x}+\left(\frac{3\sqrt{5}}{10}\right)^2+8-\left(\frac{3\sqrt{5}}{10}\right)^2\)

\(=\left(\sqrt{5x}-\frac{3\sqrt{5}}{10}\right)^2+\frac{151}{20}\)

\(=\sqrt{5}.\left(\sqrt{x}-\frac{3}{10}\right)^2+\frac{151}{20}\ge\frac{151}{20}\)(do \(\left(\sqrt{x}-\frac{3}{10}\right)^2\ge0\) )

\(\Rightarrow5x-3\sqrt{x}+8\ge\frac{151}{20}\)

\(\Rightarrow\frac{1}{5x-3\sqrt{x}+8}\le\frac{20}{151}\)

Mặt khác \(A=\frac{1}{5x-3\sqrt{x}+8}\)

\(\Rightarrow A\le\frac{20}{151}\)

Dấu ''='' xảy ra khi và chỉ khi \(\sqrt{x}=\frac{3}{10}\) hay \(x=\frac{9}{100}\)

Vậy Max A = \(\frac{20}{151}\)\(\Leftrightarrow\)\(x=\frac{9}{100}\)

\(A=\frac{1}{5x-3\sqrt{x}+8}\left(ĐKXĐ:x\ge0\right)\)Dễ dàng cm A>0

Đặt \(\sqrt{x}=t\)(\(t\ge0\))

Khi đó ta viết lại A dưới dạng \(A=\frac{1}{5t^2-3t+8}\)

\(\Leftrightarrow5t^2A-3t.A+8A-1=0\)

\(\Delta=9A^2-4.5A\left(8A-1\right)=9A^2-160A^2+20A=-151A^2+20A\ge0\)

\(\Leftrightarrow151A^2-20A\le0\)

\(\Leftrightarrow A\left(151A-20\right)\le0\)

\(\Leftrightarrow A\le\frac{20}{151}\)(Do A>0)

Vậy MAXA=20/151.Dấu "=" xảy ra khi

\(\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}A< 0\\111A-20\ge0\end{cases}}\\\hept{\begin{cases}A\ge0\\111A-20\le0\end{cases}}\end{cases}\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}A< 0\\A\ge\frac{20}{111}\end{cases}}\\\hept{\begin{cases}A\ge0\\A\le\frac{20}{111}\end{cases}}\end{cases}\Rightarrow}}A\le\frac{20}{111}\)

A = \(\frac{3x}{2}+\frac{2}{x-1}=3.\frac{x-1}{2}+\frac{2}{x-1}+\frac{3}{2}\)\(\ge2\sqrt{3}+\frac{3}{2}\)

\(\Rightarrow\)min A = \(2\sqrt{3}+\frac{3}{2}\Leftrightarrow x=\frac{2}{\sqrt{3}}+1\)(thỏa mãn)

B = \(x+\frac{3}{3x-1}=\frac{1}{3}\left(3x-1+\frac{9}{3x-1}+1\right)\)\(\ge\frac{1}{3}\left(2\sqrt{9}+1\right)=\frac{7}{3}\)

\(\Rightarrow\)min B = \(\frac{7}{3}\Leftrightarrow x=\frac{4}{3}\)

\(A\) \(=\) \(3x^2\left(8-x^2\right)\le3\frac{\left(x^2+8-x^2\right)^2}{4}=48\)

\(\Rightarrow\) maxA = 48 \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)(thỏa mãn)

\(B=\) \(4x\left(8-5x\right)\)\(=\frac{4}{5}.5x\left(8-5x\right)\le\frac{4}{5}.\frac{\left(5x+8-5x\right)^2}{4}=\frac{64}{5}\)

\(\Rightarrow\)max B = \(\frac{64}{5}\Leftrightarrow x=\frac{4}{5}\)(thỏa mãn)

C = \(4\left(x-1\right)\left(8-5x\right)=\frac{4}{5}.\left(5x-5\right)\left(8-5x\right)\)\(\le\frac{4}{5}.\frac{\left(5x-5+8-5x\right)^2}{4}=\frac{9}{5}\)

\(\Rightarrow\)max C = \(\frac{9}{5}\)\(\Leftrightarrow x=\frac{13}{10}\)(thỏa mãn)

D = \(x\left(3-\sqrt{3}\right)\)(quá dễ rồi)

\(A=\dfrac{\sqrt{x-9}}{5x}\left(ĐKx\ge9\right)\)

A'=\(\dfrac{\dfrac{5x}{2\sqrt{x-9}}-5\sqrt{x-9}}{\left(5x^2\right)}\)

\(A'=0\rightarrow5x=10\left(x-9\right)\)

            \(\rightarrow x=18\)

\(MaxA=\dfrac{1}{30}\) khi \(x=18\)

\(A=\dfrac{2.3\sqrt{x-9}}{30x}\le\dfrac{3^2+x-9}{30x}=\dfrac{1}{30}\)

\(A_{max}=\dfrac{1}{30}\) khi \(\sqrt{x-9}=3\Leftrightarrow x=18\)

a,\(\frac{x}{\sqrt{x}+1}=\frac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\frac{1}{\sqrt{x}+1}\)

\(=\left(\sqrt{x}-1\right)+\frac{1}{\sqrt{x}-1}+2\ge2.\sqrt{\left(\sqrt{x}-1\right).\frac{1}{\sqrt{x}-1}+2}\ge4\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-1=\frac{1}{\sqrt{x}-1}\)

\(\Leftrightarrow\sqrt{x}-1=1\)

\(\Leftrightarrow\sqrt{x}=2\)

\(\Leftrightarrow x=4\left(t/m\right)\)

D_min = 4  <=> x=4

b,\(\frac{\sqrt{x-9}}{5x}\) 

\(\sqrt{x-9}=\sqrt{\frac{\left(x-9\right).9}{9}}=\frac{1}{3}.\sqrt{\left(x-9\right).9}\le\frac{1}{3}.\frac{x-9+9}{2}=\frac{x}{2}\)

\(\Rightarrow D\le\frac{x}{\frac{6}{5x}}=\frac{x}{30x}=\frac{1}{30}\)

Dấu "=" xảy ra \(\Leftrightarrow x-9=9\Leftrightarrow x=18\)

D_max=\(\frac{1}{30}\Leftrightarrow x=18\)

P/s : ko chắc lắm 

\(a)\)\(P=\frac{x}{\sqrt{x}+1}=\frac{x+2\sqrt{x}+1}{\sqrt{x}+1}-\frac{2\sqrt{x}+2}{\sqrt{x}+1}+\frac{1}{\sqrt{x}+1}\)

\(P=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\frac{1}{\sqrt{x}+1}\)

\(P=\sqrt{x}+1+\frac{1}{\sqrt{x}+1}-2\ge2\sqrt{\left(\sqrt{x}+1\right).\frac{1}{\sqrt{x}+1}}-2=2-2=0\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\sqrt{x}+1=\frac{1}{\sqrt{x}+1}\)\(\Leftrightarrow\)\(x=0\)

... 

ĐKXĐ : \(x\ne0\)

\(b)\)\(D=\frac{\sqrt{x-9}}{5x}\ge\frac{0}{5x}=0\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\sqrt{x-9}=0\)\(\Leftrightarrow\)\(x=9\)

...

\(P=\sqrt{\left(x+2\right)\left(2x+1\right)}+2\sqrt{x+3}-2x\)

\(P\le\dfrac{1}{2}\left(x+2+2x+1\right)+\dfrac{1}{2}\left(4+x+3\right)-2x=5\)

\(P_{max}=5\) khi \(x=1\)