a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{9;4\right\}\end{matrix}\right.\)

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

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

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

a ĐKXĐ: \(x>0;x\ne4\)

\(\Rightarrow P=\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right):\left(\dfrac{2}{x-4}+\dfrac{1}{\sqrt{x}+2}\right)=\left(\dfrac{\sqrt{x}-2-\sqrt{x}}{\sqrt{x}+2}\right):\left(\dfrac{2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}+2}\right)=\dfrac{-2}{\sqrt{x}+2}:\left(\dfrac{2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)=-\dfrac{2}{\sqrt{x}+2}:\dfrac{\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=-\dfrac{2\left(\sqrt{x-2}\right)}{\sqrt{x}}=\dfrac{4-2\sqrt{x}}{\sqrt{x}}\)  b. Vì P và x cùng dấu \(\Rightarrow P>0\Rightarrow\dfrac{4-2\sqrt{x}}{\sqrt{x}}>0\Rightarrow4-2\sqrt{x}>0\) (vì \(\sqrt{x}>0\) ) \(\Rightarrow-2\sqrt{x}>-4\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\) kết  hợp với điều kiện

\(\Rightarrow0< x< 4\)

Lời giải:ĐK: $x>0; x\neq 1$;

\(P=\left[\frac{(\sqrt{x}-1)(x+\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)}-\frac{(\sqrt{x}+1)(x-\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+1)}\right]:\frac{2(\sqrt{x}-1)^2}{(\sqrt{x}-1)(\sqrt{x}+1)}\)

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

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

b) 

\(P=\frac{\sqrt{x}+1}{\sqrt{x}-1}=1+\frac{2}{\sqrt{x}-1}\). Để $P$ nguyên thì $\sqrt{x}-1$ là ước nguyên của $2$

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

$\Rightarrow x\in\left\{0; 2; 9\right\}$

Kết hợp với ĐKXĐ suy ra $x\in\left\{2;9\right\}$

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

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

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

b: P=1/4

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

=>\(4\left(\sqrt{x}-2\right)=3\sqrt{x}\)

=>\(4\sqrt{x}-8-3\sqrt{x}=0\)

=>\(\sqrt{x}=8\)

=>x=64

c: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{\sqrt{4+2\sqrt{3}}-2}{3\cdot\sqrt{4+2\sqrt{3}}}\)

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

1: Khi x=9 thì \(A=\dfrac{3+1}{3-1}=\dfrac{4}{2}=2\)

2:

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

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

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

b: \(2P=2\sqrt{x}+5\)

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

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\dfrac{5}{2}=\dfrac{2\sqrt{x}+5}{2}\)

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

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

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

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

=>x=1/4