ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)

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

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

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

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

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

\(ĐặtP=\dfrac{A}{B}\)

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

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

Để P<1 thì P-1<0

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

=>\(\sqrt{x}-2< 0\)

=>\(\sqrt{x}< 2\)

=>0<=x<4

mà x nguyên

nên \(x\in\left\{0;1;2;3\right\}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >25\end{matrix}\right.\)

\(A>B\left(2\sqrt{x}+5\right)\)

=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}-5}>=\dfrac{2\sqrt{x}+5}{\sqrt{x}-5}\)

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

=>\(\dfrac{-\sqrt{x}-3}{\sqrt{x}-5}>=0\)

=>\(\sqrt{x}-5< 0\)

=>\(\sqrt{x}< 5\)

=>0<=x<25

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 \(\sqrt{x}+1⋮\sqrt{x}-3\)

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

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

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

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

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

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

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

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

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

\(\Rightarrow A=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)

b) \(x=9\Rightarrow A=\dfrac{3}{3+1}=\dfrac{3}{4}\)

\(x=7-4\sqrt{3}\Rightarrow A=\dfrac{\sqrt{7-4\sqrt{3}}}{\sqrt{7-4\sqrt{3}}+1}=\dfrac{\sqrt{7-2\sqrt{12}}}{\sqrt{7-2\sqrt{12}}+1}=\dfrac{\sqrt{4-2\sqrt{3}\sqrt{4}+3}}{\sqrt{4-2\sqrt{3}\sqrt{4}+3}+1}=\dfrac{2-\sqrt{3}}{2-\sqrt{3}+1}=\dfrac{2-\sqrt{3}}{3-\sqrt{3}}=\dfrac{\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}=\dfrac{3-\sqrt{3}}{6}\)

a, ĐK: \(x\ge0,x\ne1\)

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

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

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

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

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

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

b, \(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)

Khi đó: 

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

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

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

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

c, \(A=\dfrac{1}{2}\)

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

\(\Leftrightarrow4\sqrt{x}-2=\sqrt{x}+1\)

\(\Leftrightarrow3\sqrt{x}=3\)

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

Vậy không tồn tại giá trị x thỏa mãn \(A=\dfrac{1}{2}\).

a: \(A=\dfrac{2x-6\sqrt{x}+\sqrt{x}-3-2x+4\sqrt{x}+\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{3x-3\sqrt{x}-\sqrt{x}-4}{\sqrt{x}-1}\)

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

\(=\dfrac{1}{\sqrt{x}-2}\cdot\dfrac{3x-6\sqrt{x}+2\sqrt{x}-4}{\sqrt{x}-1}=\dfrac{3\sqrt{x}+2}{\sqrt{x}-1}\)

b: Để A<2 thì \(\dfrac{3\sqrt{x}+2-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)}< 0\)

=>x<1

=>x<1

Với \(x\ge0;x\ne4\) có:

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

a

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

b

\(P^2=P+2\\ \Leftrightarrow P^2-P-2=0\\ \Leftrightarrow P^2-2P+P-2=0\\ \Leftrightarrow P\left(P-2\right)+\left(P-2\right)=0\\ \Leftrightarrow\left(P-2\right)\left(P+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}P=2\\P=-1\end{matrix}\right.\)

Với P = 2 có:

\(\dfrac{4\sqrt{x}+1}{\sqrt{x}+1}=2\\ \Leftrightarrow2\left(\sqrt{x}+1\right)=4\sqrt{x}+1\\ \Leftrightarrow2\sqrt{x}+2-4\sqrt{x}-1=0\\\Leftrightarrow -2\sqrt{x}+1=0\\\Leftrightarrow-2\sqrt{x}=-1\\\Leftrightarrow \sqrt{x}=\dfrac{1}{2}\\ \Leftrightarrow x=\dfrac{1}{4} \)

Với P = -1 có:

\(\dfrac{4\sqrt{x}+1}{\sqrt{x}+1}=-1\\ \Leftrightarrow-\sqrt{x}-1-4\sqrt{x}-1=0\\ \Leftrightarrow-5\sqrt{x}=2\\ \Leftrightarrow\sqrt{x}=-\dfrac{2}{5}\left(loại\right)\)

Vậy để \(P^2=P+2\) thì \(x=\dfrac{1}{4}\)

a: P=A:B

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

\(=\dfrac{4\sqrt{x}+1}{\sqrt{x}+1}\)

b: P^2=P+2

=>P^2-P-2=0

=>(P-2)(P+1)=0

=>P=2(nhận) hoặc P=-1(loại)

=>\(4\sqrt{x}+1=2\sqrt{x}+2\)

=>2căn x=1

=>x=1/4