Hướng dẫn giải:

Ta có: \((4xy + 7y)(2x - 1)\left( {{e^{2xy}} - {e^{4x + y + 7}}} \right) = \left[ {2x(2 - y) + y + 7} \right]{e^y}\)

\(\left. { \Leftrightarrow (4x + 7)(2xy - y)\left( {{e^{2xy - y}} - {e^{4x + 7}}} \right) = 2x(2 - y) + y + 7{\rm{ (v\`i  }}{e^y} > 0\,\,\forall y} \right)\)

\( \Leftrightarrow (4x + 7)(2xy - y).\left( {{e^{2xy - y}} - {e^{4x + 7}}} \right) = (4x + 7) - (2xy - y)\)

\( \Leftrightarrow (4x + 7).\left[ {(2xy - y){e^{2xy - y}} - 1} \right] = (2xy - y)\left[ {(4x + 7){e^{4x + 7}} - 1} \right]\)

\( \Rightarrow \frac{{(2xy - y){e^{2xy - y}} - 1}}{{2xy - y}} = \frac{{(4x + 7){e^{4x + 7}} - 1}}{{4x + 7}}\) với \(x \ne  - \frac{7}{4};x \ne \frac{1}{2};y \ne 0\).

\( \Leftrightarrow {e^{2xy - y}} - \frac{1}{{2xy - y}} = {e^{4x + 7}} - \frac{1}{{4x + 7}}\)

Xét\(f(t) = {e^t} - \frac{1}{t}\,\,(t \ne 0)\)

\( \Rightarrow f'(t) = {e^t} + \frac{1}{{{t^2}}} > 0\,\,\forall t \ne 0\)

⇒ f(t) đồng biến trên các khoảng xác định

⇒ f(2xy − y) = f(4x + 7)

TH1: (2xy − y). (4x + 7) < 0

Giả sử 2xy − y < 0 và 4x + 7 > 0. Do x, y ∈ Z nên 2xy − y ≤ −1 và 4x + 7 ≥ 1.

\( \Rightarrow \left\{ \begin{array}{l}f(2xy - y) \le f( - 1) = 1 - \frac{1}{e} < 1\\f(4x + 7) \ge f(1) = e - 1 > 1\end{array} \right.\).Do đó, \(f(2xy - y) \ne f(4x + 7)\).

TH2: (2xy − y).(4x + 7) > 0

f(2xy − y) = f(4x + 7)

⇔2xy − y = 4x + 7

\( \Rightarrow y = \frac{{4x + 7}}{{2x - 1}} = 2 + \frac{9}{{2x - 1}}\)​.

Theo bài, \(y \in \mathbb{Z}\) nên 2x − 1 ∈ {±1; ±3; ±9} ⇒ x ∈ {−4;−1;0;1;2;5}.

 Chọn C