\(P=\dfrac{x^4+x^3-3x-1}{x^2+x+1}=\dfrac{\left(x^2-1\right)\left(x^2+x+1\right)-2x}{x^2+x+1}=x^2-1-\dfrac{2x}{x^2+x+1}\)

Vì x \(\in Z\) nên để P \(\in Z\) thì : \(\dfrac{x}{x^2+x+1}\in Z\) 

Đặt \(A=\dfrac{x}{x^2+x+1}\) . Với x = 0 ; ta có : \(P=-1\in Z\)

Với x khác 0 ; ta có : \(A=\dfrac{1}{x+\dfrac{1}{x}+1}\)

Nếu x > 0 ; ta có : \(0< A\le\dfrac{1}{3}\) ( vì \(x+\dfrac{1}{x}\ge2\) )  => Ko tồn tại g/t nguyên của A (L) 

Nếu x < 0 ; ta có : \(x+\dfrac{1}{x}\le-2\)  \(\Rightarrow x+\dfrac{1}{x}+1\le-1\) 

Suy ra : \(0>A\ge\dfrac{1}{-1}=-1\)  \(\Rightarrow A=-1\) 

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

x = -1 ; ta có : P = 2 \(\in Z\) (t/m) 

Vậy ... 

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

Lời giải:

$5A+B=\frac{5\sqrt{x}+1}{2\sqrt{x}+1}$

$2(5A+B)=\frac{10\sqrt{x}+2}{2\sqrt{x}+1}=\frac{5(2\sqrt{x}+1)-3}{2\sqrt{x}+1}=5-\frac{3}{2\sqrt{x}+1}$

$5A+B$ nguyên 

$\Rightarrow 2(5A+B)$ nguyên 

$\Leftrightarrow 5-\frac{3}{2\sqrt{x}+1}$ nguyên 

$\Leftrightarrow \frac{3}{2\sqrt{x}+1}$ nguyên 

Ta thấy: $\frac{3}{2\sqrt{x}+1}\leq 3$ với mọi $x\geq 0$ và $\frac{3}{2\sqrt{x}+1}>0$ với mọi $x\geq 0$

Do đó $\frac{3}{2\sqrt{x}+1}$ nguyên thì nhận các giá trị $1,2,3$

$\Leftrightarrow x=0; \frac{1}{16}; 1$