tìm x thuộc z để x^2 - x -2/x^2 - x +2 thuộc Z

Đáp án:

$x\in \{-1;0;1;2\}.$

Giải thích các bước giải:

$\dfrac{x^2 - x -2}{x^2 - x +2} \in \mathbb{Z}\\ \Leftrightarrow \dfrac{x^2 - x +2-4}{x^2 - x +2} \in \mathbb{Z}\\ \Leftrightarrow 1-\dfrac{4}{x^2 - x +2} \in \mathbb{Z}\\ \Rightarrow \dfrac{4}{x^2 - x +2} \in \mathbb{Z}\\ x \in \mathbb{Z}, \dfrac{4}{x^2 - x +2} \in \mathbb{Z}\\ \Rightarrow 4 \ \vdots \ (x^2 - x +2)\\ Ư(4)=\{\pm 1; \pm 2; \pm 4\}\\ \Rightarrow \left[\begin{array}{l} x^2 - x +2=-4 \\ x^2 - x +2=-2 \\ x^2 - x +2=-1 \\ x^2 - x +2=1 \\ x^2 - x +2=2 \\ x^2 - x +2=4\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x^2 - x +6=0 \\ x^2 - x +4=0 \\ x^2 - x +3= 0\\ x^2 - x +1=0 \\ x^2 - x=0  \\ x^2 - x -2=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x^2 - x +\dfrac{1}{4}+\dfrac{23}{4}=0 \\ x^2 - x +\dfrac{1}{4}+\dfrac{15}{4}=0 \\ x^2 - x +\dfrac{1}{4}+\dfrac{11}{4}= 0\\ x^2 - x +\dfrac{1}{4}+\dfrac{3}{4}=0 \\ x(x - 1)=0  \\ x^2 -2 x +x-2=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left(x-\dfrac{1}{2}\right)^2+\dfrac{23}{4}=0 (\text{Vô nghiệm}) \\ \left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0 (\text{Vô nghiệm}) \\ \left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}=0 (\text{Vô nghiệm}) \\ \left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0 (\text{Vô nghiệm})\\ x(x - 1)=0  \\ x^2 -2 x +x-2=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l}  x=0 \\ x=1 \\ x(x -2 ) +x-2=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l}  x=0 \\ x=1 \\ (x+1)(x -2 ) =0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l}  x=0 \\ x=1 \\ x=-1 \\ x=2\end{array} \right..$

`(x^2-x-2)/(x^2-x+2)`

`=(x^2-x+2-4)/(x^2-x+2)`

`= 1-4/(x^2-x+2)`

Để biểu thức nguyên

`->4\vdots x^2-x+2`

`->x^2-x+2\in Ư(4)={1;-1;2;-2;4;-4}`

Do `x^2-x+2=(x-1/2)^2+1,75>= 1,75∀x`

`->x^2-x+2\in {2;4}`

$\bullet$ `x^2-x+2=2`

`->x^2-x=0`

`->x(x-1)=0`

`->x=0` hoặc `x=1` (Tm)

$\bullet$ `x^2-x+2=4`

`->x^2-x-2=0`

`->x^2-2x+x-2=0`

`->x(x-2)+(x-2)=0`

`->(x+1)(x-2)=0`

`->x=-1` hoặc `x=2` (Tm)

Vậy `x\in {0;±1;2}` để biểu thức nguyên