Cho $\Large 0\leq x, y\leq 1$ thỏa mãn $\Large \dfrac{2018^{1-x}}{2018

Cho $\Large 0\leq x, y\leq 1$ thỏa mãn $\Large \dfrac{2018^{1-x}}{2018^y}=\dfrac{x^2+2019}{y^2-2y+2020}$. Gọi M, m lần lượt là GTLN, GTNN của biểu thức $\Large P=(4x^2+3y)(4y^2+3x)+25xy$, khi đó M + m bằng bao nhiêu?

• A.

  A. $\Large \dfrac{391}{16}$

   
• B.

  B. $\Large \dfrac{383}{16}$

• C.

  C. $\Large \dfrac{136}{3}$

• D.

  D. $\Large \dfrac{25}{2}$

Chọn A

$\Large \dfrac{2018^{1-x}}{2018^y}=\dfrac{x^2+2019}{y^2-2y+2020}\Leftrightarrow \dfrac{2018^{1-x}}{2018^y}=\dfrac{x^2+2019}{(y-1)^2+2019}\Leftrightarrow \dfrac{2018^{1-y}}{2018^x}=\dfrac{x^2+2019}{(y-1)^2+2019}$

$\Large \Leftrightarrow 2018^x(x^2+2019)=2018^{1-y}\left((1-y)^2+2019\right)$ (1)

Xét hàm số:

$\Large y=f(t)=2018^t(t^2+2019), t\in [0; 1]$

$\Large y'=f'(t)=2018^t.\ln2018 (t^2+2019)+2t. 2018^t\\\Large =2018^t.(t^2\ln2018+2t+2019.\ln 2018)>0, \forall t\in [0; 1]$

$\Large \Rightarrow$ hàm số đồng biến trên đoạn [0; 1]

Phương trình (1) trở thành $\Large f(x)=f(1-y)\Leftrightarrow x=1-y\Leftrightarrow x+y=1$

Ta có:

$\Large P=(4x^2+3y)(4y^2+3x)+25xy=16x^2y^2+12x^3+12y^3+9xy+25xy$

$\Large =16x^2y^2+12\left[(x+y)^3-3xy(x+y)\right]+34xy=16x^2y^2+12-36xy+34xy=16x^2y^2-2xy+12$

Với $\Large x,y\in [0; 1], x+y =1\Rightarrow 0\leq xy\leq \dfrac{1}{4}$

Đặt $\Large xy=z, z\in\left[0; \dfrac{1}{4}\right]$, ta có: $\Large P=g(z)=16z^2-2z+12, g'(z)=32z-2=0\Rightarrow z=\dfrac{1}{16}$

Mà $\Large g(0)=12, g\left(\dfrac{1}{16}\right)=\dfrac{191}{16}, g\left(\dfrac{1}{4}\right)=\dfrac{25}{2}\Rightarrow M=\dfrac{25}{2}, m=\dfrac{191}{16}\Rightarrow M+m=\dfrac{391}{16}$