a) đk: \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)

b) Ta có:

\(P=\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{3x-8\sqrt{x}+27}{9-x}\)

\(P=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)+2\sqrt{x}\cdot\left(\sqrt{x}-3\right)-3x+8\sqrt{x}-27}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{x+5\sqrt{x}+6+2x-6\sqrt{x}-3x+8\sqrt{x}-27}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{7\sqrt{x}-21}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{7\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{7}{\sqrt{x}+3}\)

c) Nếu x không là số chính phương => P vô tỉ (loại)

=> x là số chính phương khi đó để P nguyên thì:

\(\left(\sqrt{x}+3\right)\inƯ\left(7\right)\) , mà \(\sqrt{x}+3\ge3\left(\forall x\ge0\right)\)

\(\Rightarrow\sqrt{x}+3=7\Leftrightarrow\sqrt{x}=4\Rightarrow x=16\)

Vậy x = 16 thì P nguyên

a) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}=1+\dfrac{4}{\sqrt{x}-2}\)

Để A nguyên thì 4 ⋮ √x - 2

\(\Rightarrow\sqrt{x}-2\inƯ\left(4\right)\)

\(\Rightarrow\sqrt{x}-2\in\left\{1;-1;2;-2;4;-4\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{3;1;4;0;6;-2\right\}\)

Mà x \(\sqrt{x}\ge0\)

=> x thuộc {9; 1; 16; 0; 36}

b) 

a) ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\\x\ne9\end{cases}}\)

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

\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}+3}{2\left(\sqrt{x}-1\right)}=\frac{-3\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=-\frac{3}{2\left(\sqrt{x}-3\right)}\)c) Để P nguyên thì \(2\left(\sqrt{x}-3\right)\in\left\{-3;-1;1;3\right\}\)=> x thuộc rỗng.

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

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

\(A=\frac{\sqrt{x}-2-\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\frac{-\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}}\)

\(A=\frac{4}{\sqrt{x}+2}\)

b, \(A=\frac{4}{\sqrt{x}+2}=\frac{2}{3}\)

=> 2cawn x + 4 = 12

=> 2.căn x = 8

=> căn x = 4

=> x = 16 (thỏa mãn)

c, có A = 4/ căn x + 2 và B  = 1/căn x - 2

=> A.B = 4/x - 4 

mà AB nguyên

=> 4 ⋮ x - 4

=> x - 4 thuộc Ư(4) 

=> x - 4 thuộc {-1;1;-2;2;-4;4}

=> x thuộc {3;5;2;6;0;8} mà x > 0 và x khác 4

=> x thuộc {3;5;2;6;8}

d, giống c thôi

\(A=\frac{\sqrt{x}+2}{\sqrt{x}-5}=\frac{\sqrt{x}-5+7}{\sqrt{x}-5}=1+\frac{7}{\sqrt{x}-5}\)

Để A là số nguyên thì \(\frac{7}{\sqrt{x}-5}\) là số nguyên

\(\Rightarrow\sqrt{x}-5\in\left\{1;7;-1;-7\right\}\)

Auto làm nốt

Chỉ làm thử thôi nhé-.-

\(B=\left(\sqrt{x+2-4\sqrt{x-2}}+\sqrt{x+2+4\sqrt{x-2}}\right):\sqrt{\frac{4}{x^2}-\frac{4}{x}+1}\left(đk:x\ge2\right)\)

\(=\left(\sqrt{x-2-2\sqrt{x-2}.2+2^2}+\sqrt{x-2+2\sqrt{x-2}.2+2^2}\right):\sqrt{\frac{4}{x^2}-\frac{4x}{x^2}+\frac{x^2}{x^2}}\)

\(=[\left(\sqrt{\left(\sqrt{x-2}-2\right)^2}+\sqrt{\left(\sqrt{x-2}+2\right)^2}\right):\sqrt{\frac{4-4x+x^2}{x^2}}\)

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

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

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