Ta có :

\(B=\left(\frac{1}{x-4}-\frac{1}{x+4\sqrt{x}+4}\right).\frac{x+2\sqrt{x}}{\sqrt{x}}\)

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

\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\right).\left(\sqrt{x}+2\right)\)

\(=\frac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}.\left(\sqrt{x}+2\right)\)

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

a) \(P=\frac{x^2-9}{x-3}+\frac{4-4\sqrt{x}+x}{2-\sqrt{x}}+\frac{4-x}{2+\sqrt{x}}\)

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

\(x+3+2-\sqrt{x}+2-\sqrt{x}\) = \(x+7-2\sqrt{x}\)

b) Tại x = 9, ta có:

P = \(x+7-2\sqrt{x}\) = 9 + 7 - 2\(\sqrt{9}\) = 10

\(A=\left(\frac{1+\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}-\frac{1-\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}\right).\sqrt{3}\)

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

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

Để \(A=\frac{B}{6}\Leftrightarrow B=6A\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}}=-18\)

\(\Rightarrow\sqrt{x}-1=-18\sqrt{x}\Rightarrow\sqrt{x}=\frac{1}{19}\Rightarrow x=\frac{1}{361}\)

a) A xác định khi \(\left\{{}\begin{matrix}x>0\\\sqrt{x}-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\\sqrt{x}\ne3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x\ne9\end{matrix}\right.\)

b)Với \(x>0;x\ne9\), ta có:

\(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)

Để A đạt giá trị nguyên thì \(\frac{4}{\sqrt{x}-3}\) đạt giá trị nguyên

Hay\(4⋮\left(\sqrt{x}-3\right)\)

Suy ra \(\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

TH1: \(\sqrt{x}-3=\pm1\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3=1\\\sqrt{x}-3=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=16\\x=4\end{matrix}\right.\)

TH2: \(\sqrt{x}-3=\pm2\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3=2\\\sqrt{x}-3=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=5\\\sqrt{x}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=25\\x=1\end{matrix}\right.\)

TH3: \(\sqrt{x}-3=\pm4\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3=4\\\sqrt{x}-3=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=7\\\sqrt{x}=-1\left(Loại\right)\end{matrix}\right.\Rightarrow x=49\)

Vậy \(x\in\left\{1;4;16;25;49\right\}\)

Ta có:

\(P=\sqrt{\frac{15}{2}}\cdot\sqrt{\frac{10\left(a-1\right)^2}{3}}\\ =\sqrt{\frac{15}{2}\cdot\frac{10\left(a-1\right)^2}{3}}\\ =\sqrt{25\left(a-1\right)^2}\\ =5\left|a-1\right|\\ =\left[{}\begin{matrix}5\left(a-1\right)\left(a=1\right)\\5\left(1-a\right)\left(a< 1\right)\end{matrix}\right.\\ =\left[{}\begin{matrix}5a-5\\5-5a\end{matrix}\right.\)

P.s: Ko chắc lắm nha :v