Đáp án đúng: D
Giải chi tiết:Theo bài ra ta có:
\(\begin{array}{l}\,\,\,\,\,3{\left( {x + y} \right)^2} + 5{\left( {x - y} \right)^2} = 4\\ \Leftrightarrow 3{x^2} + 3{y^2} + 6xy + 5{x^2} + 5{y^2} - 10xy = 4\\ \Leftrightarrow 8\left( {{x^2} + {y^2}} \right) = 4xy + 4\\ \Leftrightarrow 2\left( {{x^2} + {y^2}} \right) = xy + 1\\ \Leftrightarrow {x^2} + {y^2} = \frac{{xy + 1}}{2}\,\,\left( 1 \right)\end{array}\)
Mặt khác ta lại có \({\left( {{x^2} - {y^2}} \right)^2} = {\left( {{x^2} + {y^2}} \right)^2} - 4{x^2}{y^2}\,\,\,\left( 2 \right)\)
Đặt \(\left\{ \begin{array}{l}{x^2} + {y^2} = u\\xy = v\end{array} \right.\), từ (1) và (2) ta có: \(\left\{ \begin{array}{l}u = \frac{1}{2}\left( {v + 1} \right)\\{\left( {{x^2} - {y^2}} \right)^2} = {u^2} - 4{v^2} = \frac{1}{4}{\left( {v + 1} \right)^2} - 4{v^2}\end{array} \right.\).
\( \Rightarrow {u^2} = \frac{1}{4}{v^2} + \frac{1}{2}v + \frac{1}{4} - 4{v^2} = - \frac{{15}}{4}{v^2} + \frac{1}{2}v + \frac{1}{4}\).
Do \(\left\{ \begin{array}{l}{x^2} + {y^2} \ge 0\\{\left( {{x^2} - {y^2}} \right)^2} \ge 0\\{x^2} + {y^2} \ge 2xy\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\frac{1}{2}\left( {v + 1} \right) \ge 0\\ - \frac{{15}}{4}{v^2} + \frac{1}{2}v + \frac{1}{4} \ge 0\\\frac{1}{2}\left( {v + 1} \right) \ge 2v\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}v \ge - 1\\ - \frac{1}{5} \le v \le \frac{1}{3}\\v \le \frac{1}{3}\end{array} \right. \Leftrightarrow - \frac{1}{5} \le v \le \frac{1}{3}\).
Thay \({x^2} + {y^2} = u,\,\,xy = v\) và \(u = \frac{1}{2}\left( {v + 1} \right),\,\,{\left( {{x^2} - {y^2}} \right)^2} = - \frac{{15}}{4}{v^2} + \frac{1}{2}v + 1\) ta có:
\(\begin{array}{l}m\left( {2v + 1} \right) = 1010{\left[ {\frac{1}{2}\left( {v + 1} \right)} \right]^2} + 1010\left[ { - \frac{{15}}{4}{v^2} + \frac{1}{2}v + \frac{1}{4}} \right]\\ \Leftrightarrow m\left( {2v + 1} \right) = 1010\left[ {\frac{1}{4}{v^2} + \frac{1}{2}v + \frac{1}{4} - \frac{{15}}{4}{v^2} + \frac{1}{2}v + \frac{1}{4}} \right]\\ \Leftrightarrow m\left( {2v + 1} \right) = 1010\left[ { - \frac{7}{2}{v^2} + v + \frac{1}{2}} \right]\\ \Leftrightarrow m = \frac{{505\left( { - 7{v^2} + 2v + 1} \right)}}{{2v + 1}}\end{array}\)
Xét hàm số \(f\left( v \right) = \frac{{ - 7{v^2} + 2v + 1}}{{2v + 1}}\) trên \(\left[ { - \frac{1}{5};\frac{1}{3}} \right]\) ta có:
\(\begin{array}{l}f'\left( v \right) = \frac{{\left( { - 14v + 2} \right)\left( {2v + 1} \right) - \left( { - 7{v^2} + 2v + 1} \right).2}}{{{{\left( {2v + 1} \right)}^2}}}\\f'\left( v \right) = \frac{{ - 28{v^2} - 14v + 4v + 2 + 14{v^2} - 4v - 2}}{{{{\left( {2v + 1} \right)}^2}}}\\f'\left( v \right) = \frac{{ - 14{v^2} - 14v}}{{{{\left( {2v + 1} \right)}^2}}}\\f'\left( v \right) = 0 \Leftrightarrow \left[ \begin{array}{l}v = 0 \in \left[ { - \frac{1}{5};\frac{1}{3}} \right]\\v = - 1 \notin \left[ { - \frac{1}{5};\frac{1}{3}} \right]\end{array} \right.\end{array}\)
Ta có: \(f\left( 0 \right) = 1,\,\,f\left( { - \frac{1}{5}} \right) = \frac{8}{{15}},\,\,f\left( {\frac{1}{3}} \right) = \frac{8}{{15}}\).
\( \Rightarrow \frac{8}{{15}} \le f\left( v \right) \le 1\,\,\forall v \in \left[ { - \frac{1}{5};\frac{1}{3}} \right] \Rightarrow \frac{{808}}{3} \le m \le 505\).
Mà \(m \in \mathbb{Z} \Rightarrow m \in \left\{ {270;271;...;505} \right\}\).
Vậy số giá trị của \(m\) thỏa mãn yêu cầu bài toán là \(505 - 270 + 1 = 236\).
Chọn D.