Đáp án đúng: A
Giải chi tiết:\(y = \frac{{{x^2} - 5x + 2}}{{2x + 2}} = \frac{{{x^2} - 5x + 2}}{{2\left( {x + 1} \right)}}\)
Để \(y\in \mathbb{Z}\) thì tử phải là số chẵn\( \Rightarrow {x^2} - 5x\,\,\, \vdots \,\,2 \Leftrightarrow x\left( {x - 5} \right)\,\, \vdots \,\,2 \Leftrightarrow \left[ \matrix{x\,\, \vdots \,\,2 \hfill \cr x - 5\,\, \vdots \,\,2 \hfill \cr} \right.\).
TH1: \(x \vdots 2 \Rightarrow \) Đặt \(x = 2x'\,\,\left( {x' \in \mathbb{Z}} \right)\) ta có:
\(\begin{array}{l}y = \frac{{4x{'^2} - 10x' + 2}}{{4x' + 2}} = \frac{{2x{'^2} - 5x + 1}}{{2x' + 1}}\\\,\,\,\, = \frac{{\left( {2x' + 1} \right)\left( {x' - 3} \right) + 4}}{{2x' + 1}} = x' - 3 + \frac{4}{{2x' + 1}}\end{array}\)
\(y\in \mathbb{Z}\Rightarrow 2x'+1\in U\left( 4 \right)=\left\{ \pm 1;\pm 2;\pm 4 \right\}\). Mà \(2x'+1\) lẻ nên \(2x'+1\in \left\{ \pm 1 \right\}\).
\(\eqalign{& 2x' + 1 = 1 \Leftrightarrow x' = 0\,\,\,\left( {tm} \right) \Rightarrow \left\{ \matrix{x = 0 \hfill \cr y = 1 \hfill \cr} \right.\,\,\left( {tm} \right) \cr & 2x' + 1 = - 1 \Leftrightarrow x' = - 1\,\,\left( {tm} \right) \Leftrightarrow \left\{ \matrix{x = - 2 \hfill \cr y = - 8 \hfill \cr} \right.\,\,\left( {ktm} \right) \cr} \)
TH2: \(x - 5\,\, \vdots \,\,2 \Rightarrow x\) lẻ \(\Rightarrow \) Đặt \(x=2x'+1\,\,\left( x'\in \mathbb{Z} \right)\), khi đó ta có:
\(\eqalign{ y = {{{{\left( {2x' + 1} \right)}^2} - 5\left( {2x' + 1} \right) + 2} \over {2\left( {2x' + 1} \right) + 2}} = {{4x{'^2} - 6x' - 2} \over {4x' + 4}} \cr \,\,\,\, = {{2x{'^2} - 3x' - 1} \over {2x' + 2}} = x - {5 \over 2} + {4 \over {2x' + 2}} = x - {5 \over 2} + {2 \over {x' + 1}} \cr}\)
Để \(y\in \mathbb{Z}\Rightarrow -\frac{5}{2}+\frac{2}{x'+1}\in \mathbb{Z}\Leftrightarrow -\frac{5}{2}+\frac{1}{\frac{x'}{2}+\frac{1}{2}}\in \mathbb{Z}\).
Do \( - {5 \over 2} + {1 \over {{{x'} \over 2} + {1 \over 2}}} \in Z\Leftrightarrow \left\{ \matrix{{{x'} \over 2} + {1 \over 2} = 2 \hfill \cr {{x'} \over 2} + {1 \over 2} = - 2 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{x' = 3 \hfill \cr x' = - 5 \hfill \cr} \right.\)
\( \Rightarrow \left\{ \matrix{ x = 7;\,\,y = 1\,\,\,\,\,\,\,\,\,\left( {tm} \right) \hfill \cr x = - 9;\,\,y = - 8\,\,\left( {ktm} \right) \hfill \cr} \right.\,\)
Vậy có 2 điểm thỏa mãn yêu cầu bài toán.
Chọn A.
