Đáp án đúng: A
Giải chi tiết:\(\begin{align} & \frac{{{2018}^{1-x}}}{{{2018}^{y}}}=\frac{{{x}^{2}}+2019}{{{y}^{2}}-2y+2020}\Leftrightarrow \frac{{{2018}^{1-x}}}{{{2018}^{y}}}=\frac{{{x}^{2}}+2019}{{{\left( y-1 \right)}^{2}}+2019}\Leftrightarrow \frac{{{2018}^{1-y}}}{{{2018}^{x}}}=\frac{{{x}^{2}}+2019}{{{\left( y-1 \right)}^{2}}+2019} \\ & \Leftrightarrow {{2018}^{x}}\left( {{x}^{2}}+2019 \right)={{2018}^{1-y}}\left( {{\left( 1-y \right)}^{2}}+2019 \right)\,\,(1) \\\end{align}\)
Xét hàm số \(y=f(t)={{2018}^{t}}.\left( {{t}^{2}}+2019 \right),\,\,t\in \left[ 0;1 \right]\)
\(y'=f'(t)={{2018}^{t}}\ln 2018.\left( {{t}^{2}}+2019 \right)+2t{{.2018}^{t}}={{2018}^{t}}.\left( {{t}^{2}}\ln 2018+2t+2019\ln 2018 \right)>0,\,\,\forall t\in \left[ 0;1 \right]\)
\(\Rightarrow \) Hàm số đồng biến trên đoạn \(\left[ 0;1 \right]\)
Phương trình (1) trở thành \(f(x)=f(1-y)\Leftrightarrow x=1-y\Leftrightarrow x+y=1\)
Ta có:
\(\begin{array}{l}P = \left( {4{x^2} + 3y} \right)\left( {4{y^2} + 3x} \right) + 25xy\\\,\,\,\,\, = 16{x^2}{y^2} + 12{x^3} + 12{y^3} + 9xy + 25xy\\\,\,\,\,\, = 16{x^2}{y^2} + 12\left( {{{\left( {x + y} \right)}^3} - 3xy\left( {x + y} \right)} \right) + 34xy\\\,\,\,\,\, = 16{x^2}{y^2} + 12 - 36xy + 34xy = 16{x^2}{y^2} - 2xy + 12\end{array}\)
Với \(x,\,y\in \left[ 0;1 \right],\,\,x+y=1\): \(0\le xy\le \frac{1}{4}\). Đặt \(xy=z,\,\,\,z\in \left[ 0;\frac{1}{4} \right]\), ta có: \(P=g(z)=16{{z}^{2}}-2z+12,\,\,g'(z)=32z-2=0\Rightarrow z=\frac{1}{16}\)
Mà \(g\left( 0 \right)=12,\,\,g\left( \frac{1}{16} \right)=\frac{191}{16},\,g\left( \frac{1}{4} \right)=\frac{25}{2}\Rightarrow M=\frac{25}{2},\,\,m=\frac{191}{16}\Rightarrow M+m=\frac{391}{16}\)
Chọn: A
