a)\(\frac{2\sqrt{x}+6}{x-9}\)=\(\frac{2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)=\(\frac{2}{\sqrt{x}-3}\)

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

     \(\frac{3\sqrt{x}-\sqrt{6}}{3x-2}\)=\(\frac{\sqrt{3}\left(\sqrt{3x}-\sqrt{2}\right)}{\left(\sqrt{3x}+\sqrt{2}\right)\left(\sqrt{3x}-\sqrt{2}\right)}\)= \(\frac{\sqrt{3}}{\sqrt{3x}+\sqrt{2}}\)

b)\(\frac{\sqrt{x}-3}{x-6\sqrt{x}+9}\)=\(\frac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)^2}\)= \(\frac{1}{\sqrt{x}-3}\)

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

    \(\frac{\sqrt{x}-2}{x+\sqrt{x}-6}\)= \(\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)=\(\frac{1}{\sqrt{x}+3}\)

c) \(\frac{2\sqrt{x}+2\sqrt{y}}{x-y}\) = \(\frac{2\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)=\(\frac{2}{\sqrt{x}-\sqrt{y}}\)

     \(\frac{x\sqrt{x}+y\sqrt{y}}{x-y}\)= \(\frac{\sqrt{x}^3+\sqrt{y}^3}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)= \(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)=\(\frac{x-\sqrt{xy}+y}{\sqrt{x}-\sqrt{y}}\)

      \(\frac{x-2\sqrt{xy}+y-3}{\sqrt{x}-\sqrt{y}+\sqrt{3}}\)= \(\frac{\left(\sqrt{x}-\sqrt{y}\right)^2-\sqrt{3}^2}{\left(\sqrt{x}-\sqrt{y}+\sqrt{3}\right)}\)=\(\frac{\left(\sqrt{x}-\sqrt{y}-\sqrt{3}\right)\left(\sqrt{x}-\sqrt{y}+\sqrt{3}\right)}{\left(\sqrt{x}-\sqrt{y}+\sqrt{3}\right)}\)=\(\sqrt{x}-\sqrt{y}-\sqrt{3}\)

HaHaHaHaHaHaHaHaHAaHa

a. điều kiện : \(\hept{\begin{cases}x\ge0\\x-1\ne0\\x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x>0\\x\ne1\end{cases}}}\)

b, ta có : \(A=\left(\frac{\sqrt{x}-2}{\left(x+2\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{x-1}\right)\times\frac{x-1}{2\sqrt{x}}\)

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

\(=-\frac{6\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\times\frac{x-1}{2\sqrt{x}}=\frac{3}{\sqrt{x}+1}\)

c. Để A nguyên thì \(\sqrt{x}+1\) là ước của 3 hay : \(\orbr{\begin{cases}\sqrt{x}+1=1\\\sqrt{x}+1=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(\text{ loại}\right)\\x=4\end{cases}}}\)

\(A=\left(\sqrt{x}-\frac{x+2\sqrt{x}+3}{\sqrt{x}+3}\right).\left(1+\frac{5}{\sqrt{x}-2}\right)=\left(\frac{x+3\sqrt{x}-\left(x+2\sqrt{x}+3\right)}{\sqrt{x}+3}\right).\left(\frac{\sqrt{x}-2+5}{\sqrt{x}-2}\right)\)

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

b.ta có \(A=1-\frac{1}{\sqrt{x}-2}>1\Leftrightarrow\frac{1}{\sqrt{x}-2}< 0\Leftrightarrow0\le x< 4\)

a. điều kiện xác định

\(\hept{\begin{cases}x\ge0\\\sqrt{x}\ne3\\\sqrt{x}\ne2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne9\\x\ne4\end{cases}}\)

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

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

c.\(A< -1\Leftrightarrow\frac{3\sqrt{x}+10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}< -1\)

\(\Leftrightarrow\frac{3\sqrt{x}+10+x-5\sqrt{x}+6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}< 0\Leftrightarrow\frac{x-2\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)+16}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}< 0\)

hay \(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)< 0\) ( Do tử số dương ) 

nên \(2< \sqrt{x}< 3\Leftrightarrow4< x< 9\)