a) \(ĐKXĐ:x\ne4;x\ne9\)

b) \(A=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)

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

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

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

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

c) Ta có: \(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\left(\sqrt{x}-3\right)+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)

\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\) (ĐK: x thuộc Z)

Vậy để A thuộc Z khi x = {2;\(\sqrt{2};\sqrt{5};\sqrt{1};\sqrt{7}\) }

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

\(=\left[\frac{x+3+\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-3}\)

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

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

c) để A>1/3 

\(\Rightarrow\frac{\sqrt{x}+3-2}{\sqrt{x}+3}>\frac{1}{3}\)

\(\Rightarrow\frac{2}{\sqrt{x}+3}>\frac{2}{3}\)

\(\Rightarrow\sqrt{x}+3>3\)

\(\Rightarrow x>0\)

ĐKCĐ: \(x\ge0;x\ne9,x\ne4\)

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

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

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

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

b, \(A\inℤ\Leftrightarrow\frac{3}{\sqrt{x}-2}\inℤ\)

Nếu x không là số chính phương thì  \(\sqrt{x}\)là số vô tỉ thì \(\sqrt{x}-2\)là số vô tỉ\(\Rightarrow A=\frac{3}{\sqrt{x}-2}\)là số vô tỉ

Nếu x là số chính phương thì \(\sqrt{x}\)là số nguyên thì \(\sqrt{x}-2\inℤ\Rightarrow\sqrt{x}-2\inƯ\left(3\right)\Rightarrow\sqrt{x}-2\in\left\{\pm1;\pm3\right\}\Rightarrow\sqrt{x}\in\left\{1;3;5\right\}\)\(\Rightarrow x\in\left\{1;9;25\right\}\)

Mà theo ĐKXĐ có x khác 9 => \(x\in\left\{1,25\right\}\)

\(\frac{\sqrt{x}+3}{\sqrt{x}+1}=1+\frac{2}{\sqrt{x}+1}\in Z\Rightarrow\frac{2}{\sqrt{x}+1}\in Z\)

giả sử \(\sqrt{x}\)là số vô tỉ=>\(\sqrt{x}+1\)là số vô tỉ 

=>\(\frac{2}{\sqrt{x}+1}\)là số vô tỉ(vô lí)

với \(\sqrt{x}\in Q\)=>\(\sqrt{x}\in Z\Rightarrow\sqrt{x}+1\in Z\)

mà \(\sqrt{x}+1\ge1\)

Vậy x=0;1 thì \(A\in Z\)

=>\(\sqrt{x}+1\in\left\{1;2\right\}\Rightarrow x\in\left\{0;1\right\}\)

Đặt \(\sqrt{x}=t\)

 => t \(\ge\) 0

\(\Rightarrow\)Để A thuộc Z thì:

\(\frac{t+3}{t+1}\in Z\)

\(=>\left(\frac{t+3}{t+1}-1\right)\in Z\)

\(\frac{2}{t+1}\in Z\)

=> \(2⋮\left(t+1\right)\Rightarrow\left(t+1\right)\inƯ\left(2\right)\)

\(\Rightarrow\left(t+1\right)\in\left\{2;-2;1;-1\right\}\)

=> \(t\in\left\{1;-3;0;-2\right\}\)

Vì \(t\ge0\)nên chỉ có t = 1; t = 0 là thoả mãn điều kiện của t

Vì \(t=\sqrt{x}\)nên :

\(x\in\left\{1;0\right\}\)

Vậy,\(x\in\left\{1;0\right\}\)