I) Đk: x > 0 và x \(\ne\)9

\(D=\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)

\(D=\frac{x+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}}\)

\(D=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)

=> \(\frac{1}{D}=\frac{\sqrt{x}+3}{\sqrt{x}+1}=\frac{\sqrt{x}+1+2}{\sqrt{x}+1}=1+\frac{2}{\sqrt{x}+1}\)

Để 1/D nguyên <=> \(\frac{2}{\sqrt{x}+1}\in Z\)

<=> \(2⋮\left(\sqrt{x}+1\right)\) <=> \(\sqrt{x}+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

Do \(x>0\) => \(\sqrt{x}+1>1\) => \(\sqrt{x}+1=2\)

<=> \(\sqrt{x}=1\) <=> x = 1 (tm)

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

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

\(E=\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

b) Với x\(\ge\)0; ta có:

\(E=\frac{8}{9}\) <=> \(\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\)

<=> \(3\sqrt{x}=2x-2\sqrt{x}+2\)

<=> \(2x-4\sqrt{x}-\sqrt{x}+2=0\)

<=> \(\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)

<=> \(\orbr{\begin{cases}x=\frac{1}{4}\left(tm\right)\\x=4\left(tm\right)\end{cases}}\)

e) Ta có: \(E=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\ge0\forall x\in R\) (vì \(x-\sqrt{x}+1=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\))

Dấu "=" xảy ra<=> x = 0

Vậy MinE = 0 <=> x = 0

Lại có: \(\frac{1}{E}=\frac{3\left(x-\sqrt{x}+1\right)}{4\sqrt{x}}=\frac{3}{4}\left(\sqrt{x}-1+\frac{1}{\sqrt{x}}\right)\ge\frac{3}{4}\left(2\sqrt{\sqrt{x}\cdot\frac{1}{\sqrt{x}}}-1\right)\)(bđt cosi)

=> \(\frac{1}{E}\ge\frac{3}{2}.\left(2-1\right)=\frac{3}{2}\)=> \(E\le\frac{2}{3}\)

Dấu "=" xảy ra<=> \(\sqrt{x}=\frac{1}{\sqrt{x}}\) <=> x = 1

Vậy MaxE = 2/3 <=> x = 1

Gọi vận tốc ô tô đi nhanh hơn là x ( km/h)  ĐK:  \(x>5\)

=> vận tốc ô tô đi chậm hơn là x-5 (km/h) 

Vì 2 ô tô đi ngược chiều nhau và gặp nhau nên ta có pt sau :

\(3\left(x-5\right)+4x=335\)

\(\Leftrightarrow x=50\)

vận tốc ô tô đi chậm hơn là 50-5=45 ( km/h) 

vậy ...

Đặt \(\hept{\begin{cases}\sqrt[3]{x+2}=a\\\sqrt[3]{5-x}=b\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a+b=1\\a^3+b^3=7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a+b=1\\\left(a+b\right)\left(a^2-ab+b^2\right)=7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a+b=1\\a^2-ab+b^2=7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a+b=1\\\left(a+b\right)^2-3ab=7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a+b=1\\1-3ab=7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a+b=1\\ab=-2\end{cases}}\)

Làm nốt

Ta có: \(\sqrt[3]{x+2}+\sqrt[3]{5-x}=1\)

<=> \(x+2+5-x+3\sqrt[3]{\left(x+2\right)\left(5-x\right)}\left[\sqrt[3]{x+2}+\sqrt[3]{5-x}\right]=1\)

<=> \(7+3\sqrt[3]{\left(x+2\right)\left(5-x\right)}.1=1\)

<=> \(3\sqrt[3]{\left(x+2\right)\left(5-x\right)}=-6\)

<=> \(\sqrt[3]{\left(x+2\right)\left(5-x\right)}=-2\)

<=>\(3x+10-x^2=-8\)

<=> \(x^2-3x-18=0\)

<=> \(x^2-6x+3x-18=0\)

<=> \(\left(x-6\right)\left(x+3\right)=0\)

<=> \(\orbr{\begin{cases}x=6\\x=-3\end{cases}}\)

Đk: \(-3\le x\le\frac{13}{4}\)

Ta có: \(2\sqrt{x+3}+\sqrt{13-4x}=x^2+4x+2\)

<=> \(2\left(\sqrt{x+3}-2\right)+\left(\sqrt{13-4x}-3\right)-x^2-4x+5=0\)

<=> \(2\cdot\frac{x+3-4}{\sqrt{x+3}+2}+\frac{13-4x-9}{\sqrt{13-4x}+3}-\left(x-1\right)\left(x+5\right)=0\)

<=> \(2\cdot\frac{x-1}{\sqrt{x+3}+2}-\frac{4x-4}{\sqrt{13-4x}+3}-\left(x-1\right)\left(x+5\right)=0\)

<=> \(\left(x-1\right)\left(\frac{2}{\sqrt{x+3}+2}-\frac{4}{\sqrt{13-4x}+3}-x-5\right)=0\)

<=> \(\orbr{\begin{cases}x=1\\\frac{2}{\sqrt{x+3}+2}-\frac{4}{\sqrt{13-4x}+3}-x-5=0\left(1\right)\end{cases}}\)

Do \(-3\le x\le\frac{13}{4}\)

=> \(\frac{2}{\sqrt{x+3}+2}\le1\); \(-\frac{4}{\sqrt{13-4x}+3}< 0\); \(-x-5< -\left(-3\right)-5=-2\)

=> \(\frac{2}{\sqrt{x+3}+2}-\frac{4}{\sqrt{13-4x}+3}-x-5< 1-2=-1< 0\)

=>  pt (1) vô nghiệm

Vậy S = {1}

Cách 1:

\(\sqrt{4x-1}+\sqrt{4x^2-1}=1\)

\(\Leftrightarrow\left(\sqrt{4x-1}-1\right)+\sqrt{4x^2-1}=0\)

\(\Leftrightarrow\frac{4x-2}{\sqrt{4x-1}+1}+\sqrt{\left(2x-1\right)\left(2x+1\right)}=0\)

\(\Leftrightarrow\sqrt{2x-1}\left(\frac{2\sqrt{2x-1}}{\sqrt{4x-1}+1}+\sqrt{2x+1}\right)=0\)

\(\Leftrightarrow\sqrt{2x-1}=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

Cách 2:

Điều kiện: \(x\ge\frac{1}{2}\)

Ta có:
\(VT=\sqrt{4x-1}+\sqrt{4x^2-1}\ge\sqrt{4.\frac{1}{2}-1}+\sqrt{4.\left(\frac{1}{2}\right)^2-1}=1=VP\)

Dấu = xảy ra khi \(x=\frac{1}{2}\)

 Đk: x \(\le\)2028

Ta có: \(\sqrt{2028-x}+\sqrt{2093-x}+\sqrt{2268-x}=29\)

<=> \(\sqrt{2028-x}-4+\sqrt{2093-x}-9+\sqrt{2268-x}-16=0\)

<=> \(\frac{2028-x-16}{\sqrt{2028-x}+4}+\frac{2093-x-81}{\sqrt{2093-x}+9}+\frac{2268-x-256}{\sqrt{2268-x}+16}=0\)

<=> \(\left(2012-x\right).\left(\frac{1}{\sqrt{2028-x}+4}+\frac{1}{\sqrt{2093-x}+9}+\frac{1}{\sqrt{2268-x}+16}\right)=0\)

<=> x = 2012 (tm)