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

\(P=\frac{1}{\sqrt{x}+1}:\left[\frac{(\sqrt{x}+3)(\sqrt{x}-3)}{(\sqrt{x}-2)(\sqrt{x}-3)}-\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-3)(\sqrt{x}-2)}+\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}-3)}\right]\)

\(=\frac{1}{\sqrt{x}+1}:\frac{x-9-(x-4)+\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}-3)}=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{(\sqrt{x}-2)(\sqrt{x}-3)}=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)

Để $P>0\Leftrightarrow \frac{\sqrt{x}-2}{\sqrt{x}+1}>0$

$\Leftrightarrow \sqrt{x}-2>0$ (do $\sqrt{x}+1>0$)

$\Leftrightarrow x>4$

Kết hợp với ĐKXĐ suy ra $x>4; x\neq 9$

a, \(P=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)

\(P=\left(\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)

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

\(P=\dfrac{1}{\sqrt{x}+1}:\left[\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)

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

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

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

\(=\dfrac{\sqrt{x}-1+\sqrt{x}}{\sqrt{x}-1}:\dfrac{x-9+x-4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{2x+\sqrt{x}-11}\)

\(=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(2x+\sqrt{x}-11\right)}\)

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

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

`a)(sqrtx-3)/(sqrtx-2)-(2sqrtx-1)/(sqrtx-1)+(x-2)/(x-3sqrtx+2)`

`=(x-4sqrtx+3-(2sqrtx-1)(sqrtx-2)+x-2)/(x-3sqrtx+2)`

`=(2x-4sqrtx+1-2x+5sqrtx-2)/(x-3sqrtx+2)`

`=(sqrtx-1)/(x-3sqrtx+2)`

`=1/(sqrtx-2)`

`b)((x+2)/(xsqrtx-1)-sqrtx/(x+sqrtx+1)+1/(1-sqrtx)):(sqrtx-1)/2`

`=((x+2-x+sqrtx-x-sqrtx-1)/(xsqrtx-1))*2/(sqrtx-1)`

`=(1-x)/(xsqrtx-1)*2/(sqrtx-1)`

`=(-2(sqrtx+1))/(x+sqrtx+1)`

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

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

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

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

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

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

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

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

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

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

a) ĐKXĐ có thêm \(x\ne4\)

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

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

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

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

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

 \(B=\left(\dfrac{x}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

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

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

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

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

\(A=\dfrac{2x+2}{\sqrt{x}}+\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\left(x>0,x\ne1\right)\)

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

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

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

\(B=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(x\ge0,x\ne4;9\right)\)

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

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

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

\(C=\left(\dfrac{x+\sqrt{x}-1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\right):\dfrac{1}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)

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

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

a) Ta có: \(\dfrac{6}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}-\dfrac{10}{\sqrt{5}}\)

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

\(=\dfrac{3}{2}\left(\sqrt{5}-1\right)+\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-2\sqrt{5}\)

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

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

=-1

Bài 1: 

a) Thay \(x=\dfrac{1}{4}\)vào B, ta được:

\(B=1:\left(\dfrac{1}{4}\cdot\dfrac{1}{2}+27\right)=1:\left(27+\dfrac{1}{8}\right)=\dfrac{8}{217}\)

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

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

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

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

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

c) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)

\(\Leftrightarrow\dfrac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)

\(\Leftrightarrow3-\sqrt{x}>0\)

\(\Leftrightarrow\sqrt{x}< 3\)

hay x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne4\end{matrix}\right.\)

a)ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

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

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

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

a) Đk: \(x>0;x\ne9;x\ne25\)

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

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

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

\(=\dfrac{3\sqrt{x}+x}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{-\sqrt{x}+5}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\sqrt{x}\left(3+\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}.\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\sqrt{x}-5}\)

\(=\dfrac{x}{\sqrt{x}-5}\)

b) Đk: \(x\ge0;x\ne1;x\ne25\)

Biểu thức

\(=\left[\dfrac{\sqrt{x}-2}{\sqrt{x}+5}+\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{x+9}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right]:\dfrac{1-\sqrt{x}}{5+\sqrt{x}}\)

\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)+\sqrt{x}\left(\sqrt{x}+5\right)-x-9}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}.\dfrac{\sqrt{x}+5}{1-\sqrt{x}}\)

\(=\dfrac{x-7\sqrt{x}+10+x+5\sqrt{x}-x-9}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}\)

\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}\)\(=\dfrac{\left(1-\sqrt{x}\right)^2}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}=\dfrac{1-\sqrt{x}}{\sqrt{x}-5}\)

Đk: \(x>0;x\ne4\)

Biểu thức:

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

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

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

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

Khoảng cách hai ta là 100 bước, em bước 100, anh lùi 1