Đáp án:
$Q\ge 2\sqrt{2020.2017}-4037$
Giải thích các bước giải:
$Q=\dfrac{2x^2-y^2+x+1}{x+2020}$
$\rightarrow Q=\dfrac{x^2+(x^2-y^2)+x+1}{x+2020}$
$\rightarrow Q=\dfrac{x^2+(x-y)(x+y)+x+1}{x+2020}$
$\rightarrow Q=\dfrac{x^2+(x-y).1+x+1}{x+2020}$
$\rightarrow Q=\dfrac{x^2+2x+1-y}{x+2020}$
$\rightarrow Q=\dfrac{x^2+2x+x}{x+2020}$
$\rightarrow Q=\dfrac{x^2+3x}{x+2020}$
$\rightarrow Q=\dfrac{x^2-2020^2+3x+3.2020+2020^2-3.2020}{x+2020}$
$\rightarrow Q=\dfrac{(x-2020)(x+2020)+3(x+2020)+2020(2020-3)}{x+2020}$
$\rightarrow Q=x-2020+3+\dfrac{2020.2017}{x+2020}$
$\rightarrow Q=x+2020+\dfrac{2020.2017}{x+2020}-4037$
$\rightarrow Q\ge 2\sqrt{(x+2020).\dfrac{2020.2017}{x+2020}}-4037$
$\rightarrow Q\ge 2\sqrt{2020.2017}-4037$
