\(\dfrac{\sqrt{x}-5}{\sqrt{x-3}}=1-\dfrac{2}{\sqrt{x}-3}=P\)

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

\(\begin{matrix}\sqrt{x}-3&-1&-2&1&2\\\sqrt{x}&-2\left(L\right)&1&4&5\\x&&1\left(tm\right)&16\left(tm\right)&25\left(tm\right)\end{matrix}\) 

Mà x nguyên lớn nhất \(\Rightarrow x=25\)

Để P là số nguyên thì

căn x-3-2 chia hết cho căn x-3

=>căn x-3 thuộc Ư(-2)

mà x nguyên lớn nhất

nên căn x-3=2

=>x=25

\(P=\dfrac{\sqrt x-5}{\sqrt x -3}=\dfrac{\sqrt x-3-2}{\sqrt x -3}=1-\dfrac{2}{\sqrt x -3}\)

Để \(P \in Z \Leftrightarrow 2\vdots \sqrt x -3 \Rightarrow \sqrt x -3 \in \text{Ư(2)={1;-1;2;-2}}\)

\(\Rightarrow \sqrt x \in \text{{4;2;5;1}} \Rightarrow x \in \text{{16;4;25;1}}\)

\(\Rightarrow x_{max}=25\)

Ta có: \(Q=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

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

\(=\dfrac{2}{x-1}\)

Để Q nguyên thì x-1=2

hay x=3

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 \(\sqrt{x}+1⋮\sqrt{x}-3\)

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

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

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

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

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

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

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

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

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

Để \(P=-\dfrac{2}{5}\) thì \(\dfrac{\sqrt{x}-1}{x+1}=\dfrac{-2}{5}\)

\(\Leftrightarrow-2x-2=5\sqrt{x}-5\)

\(\Leftrightarrow-2x-2-5\sqrt{x}+5=0\)

\(\Leftrightarrow-2x-5\sqrt{x}+3=0\)

\(\Leftrightarrow-2x-6\sqrt{x}+\sqrt{x}+3=0\)

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

\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(-2\sqrt{x}+1\right)=0\)

\(\Leftrightarrow-2\sqrt{x}+1=0\)

\(\Leftrightarrow-2\sqrt{x}=-1\)

\(\Leftrightarrow x=\dfrac{1}{4}\)(thỏa ĐK)

\(P=A:B=\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

P>3/2

=>P-3/2>0

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)

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

=>-căn x+2>0

=>-căn x>-2

=>0<x<4

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

P lớn nhất khi căn x+1=1

=>x=0

a: Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}-1\)

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

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

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