Cho

B

- Hướng dẫn giải

Đáp án B

Ta có ${2017}^{1-x-y}=\frac{x^2+2018}{y^2-2y+2019}\iff \frac{{2017}^{1-y}}{{2017}^x}=\frac{x^2+2018}{{\left(1-y\right)}^2+2018}$

${2017}^x\left(x^2+2018\right)={2017}^{1-y}\left[{\left(1-y\right)}^2+2018\right]\iff f\left(x\right)=f\left(1-y\right)$

Xét hàm số $f\left(t\right)={2017}^t\left(t^2+2018\right)=t^2{.2017}^t+{2018.2017}^t,$có

                 $f\text{'}\left(t\right)=2t{.2017}^t+t^2{.2017}^t.\ln 2017+{2018.2017}^t.\ln 2017>0;\forall t>0$

Suy ra f(t) là hàm đồng biến trên $\left(0;+\infty \right)$ mà $f\left(x\right)=f\left(1-y\right)\Rightarrow x+y=1$

Lại có $P=\left(4x^2+3y\right)\left(4y^2+3x\right)+25xy=16x^2{y}^2+12x^3+12y^3+34xy$

$16x^2{y}^2+12\left[{\left(x+y\right)}^3-3xy\left(x+y\right)\right]+34xy=16x^2{y}^2+12\left(1-3xy\right)+34xy=16x^2{y}^2-2xy+12$

Mà $1=x+y\ge 2\sqrt{xy}\iff xy\le \frac{1}{4}$ nên đặt $t=xy\in \left[0;\frac{1}{4}\right]$khi đó $P=f\left(t\right)=16t^2-2t+12$

Xét hàm số $f\left(t\right)=16t^2-2y+12$ trên $\left[0;\frac{1}{4}\right]$ ta được $\left\{\begin{array}{l}\underset{\left[0;\frac{1}{4}\right]}{\min }f\left(t\right)=f\left(\frac{1}{16}\right)=\frac{191}{16}\\ \underset{\left[0;\frac{1}{4}\right]}{\max }f\left(t\right)=f\left(\frac{1}{4}\right)=\frac{25}{2}\end{array}\right.$