a) \(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne4\\x\ne9\end{cases}}\)

\(P=\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}+2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right):\left(1-\frac{3\sqrt{x}-9}{x-9}\right)\)

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

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

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

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

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

b) Ta có: \(P=\frac{\sqrt{x}+2}{\sqrt{x}}=1+\frac{2}{\sqrt{x}}\)

Vì \(x\inℤ\)\(\Rightarrow\)Để P nguyên thì \(\frac{2}{\sqrt{x}}\inℤ\)

\(\Rightarrow2⋮\sqrt{x}\)\(\Rightarrow\sqrt{x}\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Vì \(\sqrt{x}>0\)\(\Rightarrow\sqrt{x}\in\left\{1;2\right\}\)

\(\Rightarrow x\in\left\{1;4\right\}\)

So sánh với ĐKXĐ ta thấy \(x=1\)thỏa mãn 

\(\Rightarrow P=\frac{\sqrt{1}+2}{\sqrt{1}}=\frac{1+2}{1}=3\)

Vậy \(x=1\)khi đó \(P=3\)

\(P=\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}+2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\div\left(1-\frac{3\sqrt{x}-9}{x-9}\right)\)

a) ĐK : \(\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)

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

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

\(=\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\div\left(\frac{\sqrt{x}+3}{\sqrt{x}+3}-\frac{3}{\sqrt{x}+3}\right)\)

\(=\left(\frac{9-x+x-4-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\div\left(\frac{\sqrt{x}}{\sqrt{x}+3}\right)\)

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

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

b) Ta có : \(\frac{\sqrt{x}+2}{\sqrt{x}}=1+\frac{2}{\sqrt{x}}\)

Để P nguyên => \(\frac{2}{\sqrt{x}}\)nguyên

=> \(2⋮\sqrt{x}\)

=> \(\sqrt{x}\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

=> \(\sqrt{x}\in\left\{1;2\right\}\)( vì x ≥ 0 )

=> \(x\in\left\{1;4\right\}\Rightarrow x=1\)( vì x ≠ 4 )

Vậy với x = 1 thì P có giá trị nguyên

a) ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\\x\ne9\end{cases}}\)

b) \(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x-3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}+3}\right)\)

\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}+3}{2\left(\sqrt{x}-1\right)}=\frac{-3\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=-\frac{3}{2\left(\sqrt{x}-3\right)}\)c) Để P nguyên thì \(2\left(\sqrt{x}-3\right)\in\left\{-3;-1;1;3\right\}\)=> x thuộc rỗng.

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\(P=\left(\frac{3x+3}{x-9}-\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{3-\sqrt{x}}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right).ĐKXĐ:x\ge0,x\ne9\)

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

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

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

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

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

\(b,x=20-6\sqrt{11}=11-2.3\sqrt{11}+9\)

\(=\left(\sqrt{11}-3\right)^2\)

\(P=\frac{3}{\sqrt{x}+3}=\frac{3}{\sqrt{\left(\sqrt{11}-3\right)^2}+3}=\frac{3}{\sqrt{11}-3+3}=\frac{3\sqrt{11}}{11}\)

\(c,P>\frac{1}{2}\Rightarrow\frac{3}{\sqrt{x}+3}>\frac{1}{2}\)

\(\Leftrightarrow\frac{3}{\sqrt{x}+3}-\frac{1}{2}>0\)

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

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

vì \(2\left(\sqrt{x}+3\right)>0\) (nếu x=0 =>pt vô nghiệm)

\(\Rightarrow3-\sqrt{x}>0\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\)

Kết hợp ĐKXĐ: \(0< x< 9\)

\(Q=\frac{2P\sqrt{x}}{3}=\frac{2\left(\frac{3}{\sqrt{x}+3}\right)\sqrt{x}}{3}=\frac{6\sqrt{x}}{\sqrt{x}+3}\cdot\frac{1}{3}\)

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

\(=2-\frac{6}{\sqrt{x}+3}\)

\(ĐểQ\in Z\Rightarrow\frac{6}{\sqrt{x}+3}\in Z\)

\(\Rightarrow\sqrt{x}+3\inƯ\left(6\right)=\left\{1;2;3;6;-1;-2;-3;-6\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{-2;-1;0;3\right\}\)

\(\Rightarrow x\in\left\{0;9\right\}\)

Kết hợp ĐKXĐ \(\Rightarrow x=0\)