a: \(A=\dfrac{x+15+2x-6}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x+9}{\left(x+3\right)\left(x-3\right)}=\dfrac{3}{x-3}\)

\(x+\dfrac{2}{x}\in Z\Rightarrow\left\{{}\begin{matrix}x\in Z\\\dfrac{2}{x}\in Z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\in Z\\2⋮x\end{matrix}\right.\Rightarrow x\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Để \(x+\dfrac{2}{x}=\dfrac{x^2+2}{x}\) là số nguyên thì \(2⋮x\)

\(\Leftrightarrow x\in\left\{1;-1;2;-2\right\}\)

ĐKXĐ: x<>0

Để E là số nguyên thì \(x-4⋮x\)

=>\(-4⋮x\)

=>\(x\inƯ\left(-4\right)\)

=>\(x\in\left\{1;-1;2;-2;4;-4\right\}\)

ĐKXĐ: x<>0

Để F là số nguyên  thì \(x-3⋮2x\)

=>\(2x-6⋮2x\)

=>\(2x\inƯ\left(-6\right)\)

=>\(2x\in\left\{-1;1;2;-2;3;-3;6;-6\right\}\)

mà x là số nguyên

nên \(2x\in\left\{2;-2;6;-6\right\}\)

=>\(x\in\left\{1;-1;3;-3\right\}\)

ĐKXĐ: \(x\ge0;x\ne25\)

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

Để \(A\in\mathbb{Z}\) thì: \(\dfrac{7}{\sqrt{x}-5}\) nhận giá trị nguyên

\(\Rightarrow 7\vdots\sqrt{x}-5\)

\(\Rightarrow\sqrt{x}-5\inƯ\left(7\right)\)

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

\(\Rightarrow\sqrt{x}\in\left\{6;12;4;-2\right\}\) mà \(\sqrt{x}\ge0\)

\(\Rightarrow\sqrt{x}\in\left\{4;6;12\right\}\)

\(\Rightarrow x\in\left\{16;36;144\right\}\left(tm\right)\)

Vậy \(A\in \mathbb{Z}\) khi \(x\in\left\{16;36;144\right\}\)

1.

\(a,Q=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{x+5}{x-\sqrt{x}-2}\)

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

\(b,Q\in Z\Leftrightarrow\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\in Z\)

\(\Leftrightarrow\dfrac{-\left(\sqrt{x}-2\right)-8}{\sqrt{x}-2}\in Z\\ \Leftrightarrow-1-\dfrac{8}{\sqrt{x}-2}\in Z\)

Mà \(-1\in Z\Leftrightarrow\dfrac{8}{\sqrt{x}-2}\in Z\)

\(\Leftrightarrow8⋮\sqrt{x}-2\\ \Leftrightarrow\sqrt{x}-2\inƯ\left(8\right)=\left\{-8,-4,-2,-1,1,2,4,8\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{-6;-2;0;1;3;4;6;10\right\}\)

Mà \(x\in Z\) và \(\sqrt{x}\ge0\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;1;4\right\}\\ \Leftrightarrow x\in\left\{0;1;4\right\}\)

Vậy \(x\in\left\{0;1;4\right\}\) thì \(Q\in Z\)

\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(\text{đ}k\text{x}\text{đ}:x\ge3\right)\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{2\sqrt{x}-9-\left(x-9\right)-\left(2x-4\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}-9-x+9-2x+4\sqrt{x}-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{5\sqrt{x}-3x+2}{x-5\sqrt{x}+6}\)

__

Để \(M\in Z\) thì \(x-5\sqrt{x}+6\) thuộc ước của \(5\sqrt{x}-3x+2\)

\(\Rightarrow x-5\sqrt{x}+6=-5\sqrt{x}-3x+2\\ \Leftrightarrow x-5\sqrt{x}+6+5\sqrt{x}+3x-2=0\\ \Leftrightarrow4x-4=0\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\)