Chọn A
Ta có: \(x+y+z=0\Leftrightarrow x+y=-z .\)
\(2\left(xy+yz+zx\right)+1=0\Leftrightarrow 2xy+1=-2z\left(x+y\right)=2\left(x+y\right)^{2} . \)
Do \(4xy\le \left(x+y\right)^{2} \Rightarrow 2xy\le \frac{\left(x+y\right)^{2} }{2} \Rightarrow 2\left(x+y\right)^{2} =2xy+1\le \frac{\left(x+y\right)^{2} }{2} +1\)
\(\Leftrightarrow \left(x+y\right)^{2} \le \frac{2}{3} \Leftrightarrow -\sqrt{\frac{2}{3} } \le x+y\le \sqrt{\frac{2}{3} } \Rightarrow -\sqrt{\frac{2}{3} } \le z\le \sqrt{\frac{2}{3} } \)
Mặt khác ta có: \(0=\left(x+y+z\right)^{3} =\left(x+y\right)^{3} +3\left(x+y\right)^{2} z+3\left(x+y\right)z^{2} +z^{3} =x^{3} +y^{3} +z^{3} +3xy\left(x+y\right)+3z^{3} -3z^{3} \)
\(\Rightarrow x^{3} +y^{3} +z^{3} =-3xy\left(x+y\right)=3xyz . \)
Mà: \(2\left(xy+yz+zx\right)+1=0\Rightarrow xy=-\frac{1}{2} -z\left(x+y\right)=-\frac{1}{2} +z^{2} \Rightarrow x^{3} +y^{3} +z^{3} =3\left(-\frac{1}{2} +z^{2} \right)z=3z^{3} -\frac{3}{2} z\)
\(P=\frac{25}{9} \left(x^{3} +y^{3} +z^{3} \right)+\frac{71}{6} \left(x+y\right)=\frac{25}{9} \left(3z^{3} -\frac{3}{2} z\right)-\frac{71}{6} z=\frac{25}{3} z^{3} -16z\)
Xét hàm số \(P=f\left(z\right)=\frac{25}{3} z^{3} -16z với -\sqrt{\frac{2}{3} } \le z\le \sqrt{\frac{2}{3} } \)
\(\Rightarrow f'\left(z\right)=25z^{2} -16\Rightarrow f'\left(z\right)=0\Leftrightarrow \left[\begin{array}{l} {z=\frac{4}{5} \in \left[-\sqrt{\frac{2}{3} } ;\sqrt{\frac{2}{3} } \right]} \\ {z=\frac{-4}{5} \in \left[-\sqrt{\frac{2}{3} } ;\sqrt{\frac{2}{3} } \right]} \end{array}\right. \)
Ta có:
\(f\left(-\sqrt{\frac{2}{3} } \right)=\frac{94\sqrt{6} }{27} ;f\left(-\sqrt{\frac{2}{3} } \right)=\frac{-94\sqrt{6} }{27} ;f\left(\frac{4}{5} \right)=-\frac{128}{15} ;f\left(-\frac{4}{5} \right)=\frac{128}{15} \)
Vậy\( P_{Max} =\frac{128}{5} \Leftrightarrow \left(x;y;z\right)\in \left\{\left(\frac{4-\sqrt{2} }{10} ;\frac{4+\sqrt{2} }{10} ;\frac{-4}{5} \right);\left(\frac{4+\sqrt{2} }{10} ;\frac{4-\sqrt{2} }{10} ;\frac{-4}{5} \right)\right\} \)
\(P_{Min} =-\frac{128}{5} \Leftrightarrow \left(x;y;z\right)\in \left\{\left(\frac{-4-\sqrt{2} }{10} ;\frac{-4+\sqrt{2} }{10} ;\frac{4}{5} \right);\left(\frac{-4+\sqrt{2} }{10} ;\frac{-4-\sqrt{2} }{10} ;\frac{4}{5} \right)\right\}\Rightarrow M+m=0 \)