11:
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{9;1\right\}\end{matrix}\right.\)
\(P=\left(\dfrac{x+2\sqrt{x}-7}{x-9}+\dfrac{\sqrt{x}-1}{3-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(=\left(\dfrac{x+2\sqrt{x}-7}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\right):\dfrac{\sqrt{x}-1-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+2\sqrt{x}-7-\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{-4}\)
\(=\dfrac{x+2\sqrt{x}-7-x-2\sqrt{x}+3}{\left(\sqrt{x}-3\right)}\cdot\dfrac{-\left(\sqrt{x}-1\right)}{4}\)
\(=\dfrac{-4}{-4}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\)
b: Khi \(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\) thì
\(P=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}-1}{\sqrt{\left(\sqrt{3}-1\right)^2}-3}\)
\(=\dfrac{\sqrt{3}-1-1}{\sqrt{3}-1-3}=\dfrac{\sqrt{3}-2}{\sqrt{3}-4}=\dfrac{2-\sqrt{3}}{4-\sqrt{3}}\)
\(=\dfrac{5-2\sqrt{3}}{13}\)
c: P<1
=>P-1<0
=>\(\dfrac{\sqrt{x}-1-\sqrt{x}+3}{\sqrt{x}-3}< 0\)
=>\(\dfrac{2}{\sqrt{x}-3}< 0\)
=>\(\sqrt{x}-3< 0\)
=>0<=x<9
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< =x< 9\\x< >1\end{matrix}\right.\)
d: Để P nguyên thì \(\sqrt{x}-1⋮\sqrt{x}-3\)
=>\(\sqrt{x}-3+2⋮\sqrt{x}-3\)
=>\(\sqrt{x}-3\in\left\{1;-1;2;-2\right\}\)
=>\(\sqrt{x}\in\left\{4;2;5;1\right\}\)
=>\(x\in\left\{16;4;25;1\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{4;16;25\right\}\)