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

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 ... 

Để A là số nguyên nhỏ nhất thì x+3=-1

hay x=-4

làm thế nào có đc -1 v?

\(A=\frac{x^3-4x^2+4x-10}{x-3}\)( ĐKXĐ : x ≠ 3 )

\(=\frac{x^3-3x^2-x^2+3x+x-3-7}{x-3}\)

\(=\frac{x^2\left(x-3\right)-x\left(x-3\right)+\left(x-3\right)-7}{x-3}\)

\(=\frac{\left(x-3\right)\left(x^2-x+1\right)-7}{x-3}\)

\(=\frac{\left(x-3\right)\left(x^2-x+1\right)}{x-3}-\frac{7}{x-3}\)

\(=\left(x^2-x+1\right)-\frac{7}{x-3}\)

Vì x ∈ Z nên ( x² - x + 1 ) ∈ Z

nên để A ∈ Z thì \(\frac{7}{x-3}\)∈ Z

hay ( x - 3 ) ∈ Ư(7) = { ±1 ; ±7 }

Các giá trị tm ĐKXĐ

Vậy x ∈ { ±4 ; 2 ; 10 } thì A ∈ Z

\(ĐKXĐ:x\ne3\)

\(A=\frac{x^3-4x^2+4x-10}{x-3}=\frac{x^3-3x^2-x^2+3x+x-3-7}{x-3}\)

\(=\frac{x^2\left(x-3\right)-x\left(x-3\right)+\left(x-3\right)-7}{x-3}\)

\(=\frac{\left(x-3\right)\left(x^2-x+1\right)-7}{x-3}=\left(x^2-x+1\right)-\frac{7}{x-3}\)

Vì \(x\inℤ\)\(\Rightarrow x^2-x+1\inℤ\)

\(\Rightarrow\)Để \(A\inℤ\)thì \(\frac{7}{x-3}\inℤ\)\(\Rightarrow7⋮x-3\)

\(\Rightarrow x-3\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

\(\Leftrightarrow x\in\left\{-4;2;4;10\right\}\)( thỏa mãn ĐKXĐ )

Vậy \(x\in\left\{-4;2;4;10\right\}\)

Lỗi kìa

a) \(\left(\frac{x+3}{x-2}+\frac{x+2}{3-x}+\frac{x+2}{x^2-5x+6}\right):\left(\frac{1-x}{x+1}\right)\)

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

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

= \(\left(\frac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}\right).\frac{x+1}{1-x}\)

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

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

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

b) Để A >1 \(\Leftrightarrow\frac{x+1}{\left(x-2\right)\left(1-x\right)}>1\)

\(\Leftrightarrow\frac{-\left(1-x\right)\left(3-x\right)}{\left(x-2\right)\left(1-x\right)}\)

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

\(\Rightarrow\orbr{\begin{cases}x-3\ge0\\x-2>0\end{cases}\Leftrightarrow\orbr{\begin{cases}x\ge3\\x>2\end{cases}\Leftrightarrow}x\ge3}\)

\(\Rightarrow\orbr{\begin{cases}x-3< 0\\x-2< 0\end{cases}\Leftrightarrow\orbr{\begin{cases}x< 3\\x< 2\end{cases}\Leftrightarrow}x< 2}\)

Vậy ...