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

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

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

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

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

\(=\dfrac{-3}{2\sqrt{x}+4}\)

Để \(C< -\dfrac{1}{3}\) thì \(\dfrac{-3}{2\sqrt{x}+4}+\dfrac{1}{3}< 0\)

\(\Leftrightarrow-9+2\sqrt{x}+4< 0\)

\(\Leftrightarrow\sqrt{x}< \dfrac{5}{2}\)

hay \(0\le x< \dfrac{25}{4}\)

a: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{x-9}\)

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

\(=\dfrac{-3}{\sqrt{x}-3}\)

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 \(Q\in Z\)

ĐKXĐ: x>=0; \(x\notin\left\{9;4\right\}\)\(P=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}-\dfrac{\sqrt{x}-3}{2-\sqrt{x}}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

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

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

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

\(=\dfrac{3}{\sqrt{x}-2}\)

Để P là số nguyên thì \(3⋮\sqrt{x}-2\)

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

=>\(\sqrt{x}\in\left\{3;1;5;-1\right\}\)

=>\(\sqrt{x}\in\left\{3;1;5\right\}\)

=>\(x\in\left\{9;1;25\right\}\)

Kết hợp ĐKXĐ, ta được; \(x\in\left\{1;25\right\}\)

Lời giải:
ĐKXĐ: $x\geq 0; x\neq 9; x\neq 4$

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

\(=\frac{-3(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}:\frac{9-x+x-9-(\sqrt{x}-2)^2}{(\sqrt{x}-2)(\sqrt{x}+3)}\)

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

Với $x\in\mathbb{Z}$, để $P$ nguyên thì $\sqrt{x}-2$ là ước nguyên của 3

$\Rightarrow \sqrt{x}-2\in \left\{1; -1; 3; -3\right\}$

$\Rightarrow \sqrt{x}\in \left\{3; 1; 5; -1\right\}$

$\Rightarrow x\in \left\{9; 1; 25\right\}$

Theo ĐKXĐ suy ra $x=1$ hoặc $x=25$

\(B=\dfrac{2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}}{x-4\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

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

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

Để B nguyên thì \(\sqrt{x}-3\in\left\{1;-1;5\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{4;2;8\right\}\)

hay \(x\in\left\{16;4;64\right\}\)

1: Ta có: \(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

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

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

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

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

Để \(A=-\dfrac{1}{\sqrt{x}}\) thì \(x+\sqrt{x}=-\sqrt{x}+3\)

\(\Leftrightarrow x+2\sqrt{x}-3=0\)

\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow x=1\left(nhận\right)\)

2: Để A nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)

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

\(\Leftrightarrow\sqrt{x}\in\left\{2;4;5;1;7\right\}\)

\(\Leftrightarrow x\in\left\{16;25;1;49\right\}\)

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 \(x\in\left\{16;36;144\right\}\)

a) \(\left(\dfrac{x+2\sqrt{x}-7}{x-9}+\dfrac{\sqrt{x}-1}{3-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-1}\right)\)ư

=\(\dfrac{x+2\sqrt{x}-7-\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}-1-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

=\(\dfrac{-4\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{-4\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\)

b)ta có : \(\dfrac{\sqrt{x}-1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+2}{\sqrt{x}-3}=1+\dfrac{2}{\sqrt{x}-3}\)

để P nguyên thì \(\sqrt{x}-3\inƯ\left(2\right)\Leftrightarrow\sqrt{x}-3\inƯ\left(\pm1,\pm2\right)\)

\(\Rightarrow\sqrt{x}-3=1\Leftrightarrow x=16\left(TM\right)\)

\(\sqrt{x}-3=-1\Leftrightarrow x=4\left(KTM\right)\)

\(\sqrt{x}-3=2\Leftrightarrow x=25\left(TM\right)\)

\(\sqrt{x}-3=-2\Leftrightarrow x=1\left(KTM\right)\)

vậy x\(\in\left\{16,25\right\}\)

ĐKXĐ : \(x\ge9\)

\(a,x< 50\Leftrightarrow\sqrt{x}-1< 5\sqrt{2}-1\\ M=\dfrac{\sqrt{x}-1}{2}\in Z\\ \Leftrightarrow\sqrt{x}-1\in B\left(2\right)=\left\{0;2;4;6\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{1;3;5;7\right\}\\ \Leftrightarrow x\in\left\{1;9;25;49\right\}\\ b,\Leftrightarrow\sqrt{x}-5\inƯ\left(9\right)=\left\{-3;-1;1;3;9\right\}\left(\sqrt{x}-5>-5\right)\\ \Leftrightarrow\sqrt{x}\in\left\{2;4;6;8;14\right\}\\ \Leftrightarrow x\in\left\{4;16;36;64;196\right\}\)