a: \(P=\left(\dfrac{2+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\)

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

b: Để P nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-1\)

\(\Leftrightarrow\sqrt{x}-1\in\left\{-1;1;2\right\}\)

hay \(x\in\left\{0;4;9\right\}\)

a, ĐK: \(x>0;x\ne1\)

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

\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]:\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

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

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

a: Ta có: \(P=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\)

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

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

b: Để P nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-1\)

\(\Leftrightarrow\sqrt{x}-1\in\left\{1;-1;2\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{2;0;3\right\}\)

ha \(x\in\left\{4;9\right\}\)

b, \(P\in Z\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\in Z\)

\(\Leftrightarrow\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}\in Z\)

\(\Leftrightarrow1+\dfrac{2}{\sqrt{x}-1}\in Z\)

\(\Leftrightarrow\sqrt{x}-1\inƯ_2=\left\{\pm1;\pm2\right\}\)

\(\Leftrightarrow\sqrt{x}\inƯ_2=\left\{0;2;3\right\}\)

\(\Leftrightarrow x\inƯ_2=\left\{4;9\right\}\)

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

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

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

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

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

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

b: Để Q là số nguyên thì \(2x⋮x-1\)

=>\(2x-2+2⋮x-1\)

=>\(2⋮x-1\)

=>\(x-1\in\left\{1;-1;2;-2\right\}\)

=>\(x\in\left\{2;0;3;-1\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{0;2;3\right\}\)

\(P=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)-\dfrac{x\sqrt{x}+1}{x-\sqrt{x}+1}\) đk: \(x\ge0,x\ne1\)

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

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

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

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

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

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

b)Để P<4 \(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-1}< 4\) \(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-4< 0\) \(\Leftrightarrow\dfrac{\sqrt{x}+2-4\left(\sqrt{x}-1\right)}{\sqrt{x}-1}< 0\)

\(\Leftrightarrow\dfrac{6-3\sqrt{x}}{\sqrt{x}-1}< 0\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}6-3\sqrt{x}>0\\\sqrt{x}-1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}6-3\sqrt{x}< 0\\\sqrt{x}-1>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}< 2\\\sqrt{x}< 1\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}>2\\\sqrt{x}>1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}< 1\\\sqrt{x}>2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0\le x< 1\\x>4\end{matrix}\right.\)

Vậy...

c)\(P=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\) \(=1+\dfrac{3}{\sqrt{x}-1}\)

Để P nguyên khi \(\dfrac{3}{\sqrt{x}-1}\) nguyên

\(x\in Z\)\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}\in Z\\\sqrt{x}\in I\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}-1\in Z\\\sqrt{x}-1\in I\end{matrix}\right.\)

Tại \(\sqrt{x}-1\in I\Rightarrow\dfrac{3}{\sqrt{x}-1}\notin Z\) (L)

Tại\(\sqrt{x}-1\in Z\) .Để \(\dfrac{3}{\sqrt{x}-1}\in Z\)

\(\Leftrightarrow\sqrt{x}-1\inƯ\left(3\right)=\left\{-1;1;-3;3\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;2;-2;4\right\}\) mà \(\sqrt{x}\ge0\)

\(\Rightarrow\sqrt{x}\in\left\{0;2;4\right\}\) \(\Leftrightarrow x\in\left\{0;4;16\right\}\) (tm)

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

\(=\left[\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right]\sqrt{x}\left(\sqrt{x}-1\right)=\)

\(=\left[\dfrac{x+\sqrt{x}-2\sqrt{x}-2-x+\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right]\sqrt{x}\left(\sqrt{x}-1\right)=\)

\(=\left[\dfrac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right]\sqrt{x}\left(\sqrt{x}-1\right)=\)

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

b/

\(B=-\dfrac{2\left(x-1\right)+2}{x-1}=-2+\dfrac{2}{x-1}\)

Để B nguyên

\(x-1=\left\{-1;-2;1;2\right\}\Rightarrow x=\left[0;-1;2;3\right]\)

Lời giải:

ĐK: $x\geq 0; x\neq 4; x\neq 9$

a) 

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

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

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

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

Với $x$ nguyên, để $P$ nguyên thì $\sqrt{x}-3$ phải là ước nguyên của $4$

Mà $\sqrt{x}-3\geq -3$ nên:

$\Rightarrow \sqrt{x}-3\in\left\{\pm 1;\pm 2;4\right\}$

$\Rightarrow x\in \left\{4;16;1;25;49\right\}$ (đều thỏa mãn.

a: Ta có: \(A=\left(\dfrac{3x+3}{x-9}-\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{\sqrt{x}-3}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

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

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