Chọn A
Với mọi số thực x, yta có:
\(\[\begin{array}{l} {5\left(x^{2} +y^{2} \right)+6xy+16=20(x+y)} \\ {\Leftrightarrow 5\left(x+y\right)^{2} -20(x+y)+16=4xy\le (x+y)^{2} } \\ {\Leftrightarrow 4\left(x+y\right)^{2} -20(x+y)+16\le 0} \\ {\Leftrightarrow 1\le x+y\le 4} \end{array}\] \)
Ta có:
\(\[\begin{array}{l} {M=x^{3} +y^{3} +3x^{2} (y-2)+3y^{2} (x-2)+9(x+y)-12xy-2} \\ {{\rm \; \; \; \; \; =}(x+y)^{3} -6(x+y)^{2} +9(x+y)-2.} \end{array}\] \)
Đặt\( t=x+y với t\in \left[1;4\right].\)
Khi đó: \(P=f(t)=t^{3} -6t^{2} +9t-2\) liên tục trên đoạn \(t\in \left[1;4\right].
\[f'(t)=3t^{2} -12t+9\] \)
\(\[\Rightarrow f'(t)=0\Leftrightarrow \left[\begin{array}{l} {t=1} \\ {t=3} \end{array}\right. .\] \)
Ta có: \(f\eqref{GrindEQ__1_}=2;f\eqref{GrindEQ__3_}=-2;f\eqref{GrindEQ__4_}=2\)
Vậy:\( \max P={\mathop{\max }\limits_{\left[1;4\right]}} f(t)=2 khi \left[\begin{array}{l} {t=1} \\ {t=4} \end{array}\right.\)
\(\Rightarrow \left[\begin{array}{l} {\left\{\begin{array}{l} {x+y=1} \\ {xy=\frac{1}{4} } \end{array}\right. } \\ {\left\{\begin{array}{l} {x+y=4} \\ {xy=4} \end{array}\right. } \end{array}\right. \Leftrightarrow \left(x;y\right)\in \left\{\left(\frac{1}{2} ;\frac{1}{2} \right);\left(-\frac{1}{2} ;-\frac{1}{2} \right);\left(2;2\right);\left(-2;-2\right)\right\}.\)
\( \min P={\mathop{\min }\limits_{\left[1;4\right]}} f(t)=-2 khi t=3\Rightarrow \left\{\begin{array}{l} {x+y=3} \\ {xy=\frac{1}{4} } \end{array}\right. \Leftrightarrow \left(x;y\right)\in \left\{\left(\frac{3+2\sqrt{2} }{2} ;\frac{3-2\sqrt{2} }{2} \right);\left(\frac{3-2\sqrt{2} }{2} ;\frac{3+2\sqrt{2} }{2} \right)\right\}.\)
Vậy \(M+m=2+\left(-2\right)=0.\)
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