Chọn A
Ta có: \(x^{2} -xy+y^{2} =1\Leftrightarrow x^{2} +y^{2} =1+xy\)
\( +) 1+xy=x^{2} +y^{2} =\left(x+y\right)^{2} -2{\rm x}y\ge -2{\rm x}y \ge \Rightarrow {\rm x}y\ge -\frac{1}{3} .\)
\( +) 1+xy=x^{2} +y^{2} =\left(x-y\right)^{2} +2{\rm x}y\ge 2{\rm x}y\Rightarrow {\rm x}y\le 1.\)
Suy ra: \(-\frac{1}{3} \le {\rm x}y\le 1\)
Đặt \(t=xy,\, t\in \left[-\frac{1}{3} ;1\right] .\)
Ta có:\( P=\frac{x^{4} +y^{4} +1}{x^{2} +y^{2} +1} =\frac{\left(x^{2} +y^{2} \right)^{2} -2x^{2} y^{2} +1}{\left(x^{2} +y^{2} \right)+1} =\frac{\left(1+xy\right)^{2} -2x^{2} y^{2} +1}{\left(1+xy\right)+1} .\)
Khi đó \(P=\frac{-t^{2} +2t+2}{t+2} suy ra P'\left(t\right)=\frac{-t^{2} -4t+2}{\left(t+2\right)^{2} } .\)
Cho\(P'\left(t\right)=0\Leftrightarrow \left[\begin{array}{l} {t=\sqrt{6} -2\, \in \left[-\frac{1}{3} ;1\right]} \\ {t=-\sqrt{6} -2\, \, \notin \left[-\frac{1}{3} ;1\right]} \end{array}\right. .\)
Ta có: \(P\left(1\right)=1; P\left(-\frac{1}{3} \right)=\frac{11}{15} ; P\left(\sqrt{6} -2\right)=6-2\sqrt{6} \)
Vậy \(M={\mathop{{\rm max}}\limits_{\left[-\frac{1}{3} ;1\right]}} P=6-2\sqrt{6} ; m={\mathop{\min P}\limits_{\left[-\frac{1}{3} ;1\right]}} =\frac{11}{15} . Suy ra A=M+15m=17-2\sqrt{6} .\)