a: \(x=6+2\sqrt5=\left(\sqrt5+1\right)^2\)

\(\sqrt{x}-1\)

\(=\sqrt{\left(\sqrt5+1\right)^2}-1=\sqrt5+1-1=\sqrt5\)

\(2x+2-2x\cdot\sqrt{x}\)

\(=2\left(6+2\sqrt5\right)+2-2\cdot\left(6+2\sqrt5\right)\sqrt{\left(\sqrt5+1\right)^2}\)

\(=12+4\sqrt5+2-2\left(6+2\sqrt5\right)\left(\sqrt5+1\right)\)

\(=14+4\sqrt5-2\left(6\sqrt5+6+10+2\sqrt5\right)=14+4\sqrt5-2\left(8\sqrt5+16\right)\)

\(=14+4\sqrt5-16\sqrt5-32=-18-12\sqrt5\)

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

\(=\frac{\sqrt5}{-18-12\sqrt5}=\frac{-\sqrt5}{18+12\sqrt5}=\frac{-\sqrt5}{6\left(3+2\sqrt5\right)}=\frac{-\sqrt5\left(2\sqrt5-3\right)}{6\left(2\sqrt5+3\right)\left(2\sqrt5-3\right)}\)

\(=\frac{-10+3\sqrt5}{6\cdot\left(20-9\right)}=\frac{-10+3\sqrt5}{66}\)

a: \(x=6+2\sqrt5=\left(\sqrt5+1\right)^2\)

\(\sqrt{x}-1\)

\(=\sqrt{\left(\sqrt5+1\right)^2}-1=\sqrt5+1-1=\sqrt5\)

\(2x+2-2x\cdot\sqrt{x}\)

\(=2\left(6+2\sqrt5\right)+2-2\cdot\left(6+2\sqrt5\right)\sqrt{\left(\sqrt5+1\right)^2}\)

\(=12+4\sqrt5+2-2\left(6+2\sqrt5\right)\left(\sqrt5+1\right)\)

\(=14+4\sqrt5-2\left(6\sqrt5+6+10+2\sqrt5\right)=14+4\sqrt5-2\left(8\sqrt5+16\right)\)

\(=14+4\sqrt5-16\sqrt5-32=-18-12\sqrt5\)

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

\(=\frac{\sqrt5}{-18-12\sqrt5}=\frac{-\sqrt5}{18+12\sqrt5}=\frac{-\sqrt5}{6\left(3+2\sqrt5\right)}=\frac{-\sqrt5\left(2\sqrt5-3\right)}{6\left(2\sqrt5+3\right)\left(2\sqrt5-3\right)}\)

\(=\frac{-10+3\sqrt5}{6\cdot\left(20-9\right)}=\frac{-10+3\sqrt5}{66}\)

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 \(x\in\left\{16;25;64\right\}\)

a. B = \(\dfrac{\sqrt{36}}{\sqrt{36}-3}=\dfrac{6}{6-3}=2\)

a: Thay x=36 vào B, ta được:

\(B=\dfrac{6}{6-3}=\dfrac{6}{3}=2\)

a: Thay \(x=4+2\sqrt3=\left(\sqrt3+1\right)^2\) vào A, ta được:

\(A=\frac{\sqrt{\left(\sqrt3+1\right)^2}-1}{\sqrt{\left.\left(\sqrt3+1\right)^2\right.}+2}\)

\(=\frac{\sqrt3+1-1}{\sqrt3+1+2}=\frac{\sqrt3}{3+\sqrt3}=\frac{1}{\sqrt3+1}=\frac{\sqrt3-1}{2}\)

b: \(B=\frac{\sqrt{x}+3}{\sqrt{x}+1}-\frac{5}{1-\sqrt{x}}+\frac{4}{x-1}\)

\(=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)+5\left(\sqrt{x}+1\right)+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{x+7\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+6}{\sqrt{x}-1}\)

c: P=A*B

\(=\frac{\sqrt{x}+6}{\sqrt{x}-1}\cdot\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{\sqrt{x}+6}{\sqrt{x}+2}\)

Để P nguyên thì \(\sqrt{x}+6\) ⋮\(\sqrt{x}+2\)

=>\(\sqrt{x}+2+4\) ⋮\(\sqrt{x}+2\)

=>4⋮\(\sqrt{x}+2\)

=>\(\sqrt{x}+2\in\left\lbrace2;4\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace0;2\right\rbrace\)

=>x∈{0;4}

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 \(x\in\left\{4;5\right\}\)