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

ai đó giúp toi đi aaa

ĐK: x>0, 5x-3\(\sqrt{x}\)+8≠ 0

+) 5x-3\(\sqrt{x}\)+8 <0 thì A<0

+)5x-3\(\sqrt{x}\)+8>0, ta có:

\(\frac{1}{5x-3\sqrt{x}+8}\)  lớn nhất khi và chỉ khi \(5x-3\sqrt{x}+8\)bé nhất

5x-3\(\sqrt{x}\)+8 ≥ 3/10 ∀x

⇒ Min5x-3\(\sqrt{x}\)+8=3/10

⇒ GTLN của A là  1: 3/10=10/3

Sai thì thôi :v

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

\(\Rightarrow A_{max}=\frac{20}{151}\) 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,\(\frac{x}{\sqrt{x}+1}=\frac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\frac{1}{\sqrt{x}+1}\)

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

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 \(\Leftrightarrow\)\(\sqrt{x-9}=0\)\(\Leftrightarrow\)\(x=9\)

...

Áp dụng BĐT cosi:

\(A=\sqrt{\left(2x+1\right)\left(x+2\right)}+2\sqrt{x+3}-2x\\ A\le\dfrac{2x+1+x+2}{2}+\dfrac{4+x+3}{2}-2x\\ A\le\dfrac{3x+3}{2}+\dfrac{x+7}{2}-2x=\dfrac{3x+3+x+7-4x}{2}=5\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}2x+1=x+2\\4=x+3\end{matrix}\right.\Leftrightarrow x=1\)

\(\left(1\right)< =>-3\left(x-1\right)\left(x+1\right)\left(3x^2-8x-4\right)=0=>\orbr{\begin{cases}x=1\\x=\frac{4-2\sqrt{7}}{3};\frac{4+2\sqrt{7}}{3}\end{cases}.}\)
 

\(A=3+\frac{\sqrt{5x+1}}{7x+3}=3+\frac{\sqrt{5x+1}}{\frac{7}{5}\left(5x+1\right)+\frac{8}{5}}=3+\frac{1}{\frac{1}{5}\left(7\sqrt{5x+1}+\frac{8}{\sqrt{5x+1}}\right)}\le3+\frac{1}{\frac{1}{5}2\sqrt{7.8}}=...\)

\(A=\sqrt{2x^2+5x+2}+2\sqrt{x+3}-2x\)

\(2A=2\sqrt{2x^2+5x+2}+4\sqrt{x+3}-4x\)

\(2A=2\sqrt{\left(2x+1\right)\left(x+2\right)}+4\sqrt{x+3}-4x\)

\(\le2x+1+x+2+4+x+3-4x=10\)

=>2A\(\le10\Rightarrow A\le5\)

dấu bằng xảy ra \(\Leftrightarrow2x+1=x+2\)

và x+3=4

=>x=1

maxA=5 khi x=1

Khó vậy ta ????