\(P=\dfrac{-x+5\sqrt{x}-22}{x+2\sqrt{x}-15}+\dfrac{3\sqrt{x}-1}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\)

\(=\dfrac{-x+5\sqrt{x}-22}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\dfrac{3\sqrt{x}-1}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\)

\(=\dfrac{-x+5\sqrt{x}-22+\left(3\sqrt{x}-1\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-x+5\sqrt{x}-22+3x-10\sqrt{x}+3-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)

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

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

`A=((3sqrtx+6)/(x-4)+sqrtx/(sqrtx-2)):(x-9)/(sqrtx-3)(x>=0,x ne 4,x ne 9)`

`=((3(sqrtx+2))/((sqrtx-2)(sqrtx+2))+sqrtx/(sqrtx-2)):((sqrtx-3)(sqrtx+3))/(sqrtx-3)`

`=(3/(sqrtx-2)+sqrtx/(sqrtx-2)):(sqrtx+3)`

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

`=1/(sqrtx-2)`

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

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

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

Ta có: \(A=\left(\dfrac{3\sqrt{x}+6}{x-4}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\dfrac{x-9}{\sqrt{x}-3}\)

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

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

`a)(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)(x>=0,x ne 4,x ne 9)`

`=(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)`

`=(2sqrtx-9+(sqrtx-3)(sqrtx+3)+(2sqrtx+1)(sqrtx-2))/(x-5sqrtx+6)`

`=(2sqrtx-9+x-9+2x-3sqrtx-2)/(x-5sqrtx+6)`

`=(3x-sqrtx-20)/

\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}\\ =\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3-11\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}+\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{3x+15\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

a: Khi x=25 thì \(A=\dfrac{5-2}{5-3}=\dfrac{3}{2}\)

b: P=A*B

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\left(\dfrac{6x+6\sqrt{x}-12}{x+5\sqrt{x}+4}-\dfrac{5\sqrt{x}}{\sqrt{x}+4}\right)\)

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\cdot\left(\dfrac{6x+6\sqrt{x}-12}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}-\dfrac{5\sqrt{x}}{\sqrt{x}+4}\right)\)

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\cdot\dfrac{6x+6\sqrt{x}-12-5x-5\sqrt{x}}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+\sqrt{x}-12}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

c: \(\sqrt{P}< =\dfrac{1}{2}\)

=>0<=P<=1/4

=>\(\left\{{}\begin{matrix}P>=0\\P-\dfrac{1}{4}< =0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{x}-2}{\sqrt{x}-1}>=0\\\dfrac{\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{1}{4}< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\\dfrac{4\left(\sqrt{x}-2\right)-\sqrt{x}+1}{4\left(\sqrt{x}-1\right)}< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\\dfrac{3\sqrt{x}-7}{\sqrt{x}-1}< =0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\1< \sqrt{x}< =\dfrac{7}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\1< x< \dfrac{49}{9}\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\0< =x< 1\end{matrix}\right.\\x=\dfrac{49}{9}\end{matrix}\right.\)

=>\(4< =x< =\dfrac{49}{9}\)

mà x nguyên

nên \(x\in\left\{4;5\right\}\)

Với \(x\ge0;x\ne9\)

\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{9-x}\)

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

\(=\dfrac{14\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{14\sqrt{x}-2}{x-9}\)

Câu 2: 

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

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

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

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

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

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

Câu 2 : 

Gọi : vận tốc của người đi chậm là : x (km/h) ( x > 0 ) 

Vận tốc của người đi nhanh : x + 4 (km/h) 

Vi : người đi chậm đến muộn hơn : 45 phút \(=\dfrac{3}{4}\left(h\right)\)

Khi đó : 

\(\dfrac{36}{x}-\dfrac{36}{x+4}=\dfrac{3}{4}\)

\(\Leftrightarrow\left[36\cdot\left(x+4\right)-36x\right]\cdot4=3x\cdot\left(x+4\right)\)

\(\Leftrightarrow3x^2+12x-144=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=12\left(n\right)\\x=16\left(l\right)\end{matrix}\right.\)

Câu 1: 

1) Thay x=16 vào N, ta được:

\(N=\dfrac{2\cdot\sqrt{16}+1}{3-\sqrt{16}}=\dfrac{2\cdot4+1}{3-4}=\dfrac{9}{-1}=-9\)

Vậy: Khi x=16 thì N=-9

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

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

b: Ta có: \(B=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}\)

\(=\dfrac{x-4+\sqrt{x}-8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x+\sqrt{x}-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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

c: Để B là số tự nhiên thì \(\sqrt{x}+4⋮\sqrt{x}-2\)

\(\Leftrightarrow\sqrt{x}-2\in\left\{1;2;3;6\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{3;4;5;8\right\}\)

hay \(x\in\left\{16;25;64\right\}\)