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

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

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

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

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

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

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

a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)

b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)

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

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

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

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

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

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