\(a,A=\dfrac{2\sqrt{x}-2+2\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\left(x\ge0;x\ne1;x\ne9\right)\\ A=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)

\(b,A\in Z\Leftrightarrow\dfrac{\sqrt{x}-3+5}{\sqrt{x}-3}\in Z\Leftrightarrow1+\dfrac{5}{\sqrt{x}-3}\in Z\\ \Leftrightarrow\sqrt{x}-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ Mà.x\ge0\\ \Leftrightarrow\sqrt{x}\in\left\{2;4;8\right\}\\ \Leftrightarrow x\in\left\{4;16;64\right\}\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\\x\ne1\end{matrix}\right.\)

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

b) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}=1+\dfrac{5}{\sqrt{x}-3}\in Z\)

\(\Rightarrow\sqrt{x}-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)

Kết hợp đk

\(\Rightarrow x\in\left\{4;16;64\right\}\)

\(a,A=\frac{2}{\sqrt{x}-3}+\frac{2\sqrt{x}}{x-4\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-1}\)

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

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

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

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

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

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

để A nguyên \(5⋮\sqrt{x}-3\)

lập bảng ra đc 

\(x=\left\{2\right\}\)

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: \(P=\dfrac{x-\sqrt{x}-1-\sqrt{x}+1}{x-1}\cdot\dfrac{4\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)^2}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)\cdot4\left(\sqrt{x}-2\right)}{\sqrt{x}\left(x-1\right)}=\dfrac{4}{x-1}\)

Để P nguyên dương thì x-1 thuộc {1;4;2}

=>x thuộc {2;5;3}

b: x+y+z=0

=>x=-y-z; y=-x-z; z=-x-y

\(P=\dfrac{x^2}{y^2+z^2-\left(y+z\right)^2}+\dfrac{y^2}{z^2+x^2-\left(x+z\right)^2}+\dfrac{z^2}{x^2+y^2-\left(x+y\right)^2}\)

\(=\dfrac{x^2}{-2yz}+\dfrac{y^2}{-2xz}+\dfrac{z^2}{-2xy}\)

\(=\dfrac{x^3+y^3+z^3}{2xyz}\cdot\left(-1\right)\)

\(=-\dfrac{\left(x+y\right)^3+z^3-3xy\left(x+y\right)}{2xyz}\)

\(=-\dfrac{\left(-z\right)^3+z^3-3xy\cdot\left(-z\right)}{2xyz}=-\dfrac{3}{2}\)

Đặt \(A=\dfrac{15-\sqrt{x}}{\sqrt{x}+2}=\dfrac{\dfrac{15}{2}\left(\sqrt{x}+2\right)-\dfrac{17}{2}\sqrt{x}}{\sqrt{x}+2}=\dfrac{15}{2}-\dfrac{17\sqrt{x}}{2\left(\sqrt{x}+2\right)}\le\dfrac{15}{2}\)

\(A=\dfrac{17-\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=-1+\dfrac{17}{\sqrt{x}+2}>-1\)

\(\Rightarrow-1< A\le\dfrac{15}{2}\Rightarrow A=\left\{0;1;2;3;4;5;6;7\right\}\)

\(\Rightarrow\dfrac{15-\sqrt{x}}{\sqrt{x}+2}=\left\{0;1;2;3;4;5;6;7\right\}\)

\(\Rightarrow x=\left\{225;\dfrac{169}{4};\dfrac{121}{9};\dfrac{81}{16};\dfrac{49}{25};\dfrac{25}{36};\dfrac{9}{49};\dfrac{1}{64}\right\}\)