Xét các số thực dương x, y thỏa mãn 2018 ^2 (x^2-y+1)=2x+/(x+1)^2

* Hướng dẫn giải

Đáp án B.

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

$\iff A^{{\left(x+1\right)}^2}\left(x+1\right)=A^{2x+y}\left(2x+y\right),A={2018}^2$

Xét hàm số $F\left(t\right)=A^{\text{'}}t,t>0\Rightarrow$ $f\text{'}\left(t\right)=A\text{'}+t.A\text{'}\ln A>0\Rightarrow f\left(t\right)$ đồng biến với mọi $t>0.$

Suy ra $A^{{\left(x+1\right)}^2}\left(x+1\right)=A^{2x+y}\left(2x+y\right)\iff f\left({\left(x+1\right)}^2\right)=f\left(2x+y\right)\iff {\left(x+1\right)}^2=2x+y\iff y=x^2+1.$ 

Ta có $P=2y-3x=2\left(x^2+1\right)-3x=2x^2-3x+2=2{\left(x-\frac{3}{4}\right)}^2+\frac{7}{8}\ge \frac{7}{8}\Rightarrow P_{\min }=\frac{7}{8}$