cho 2 số thực x, y dương thoả mãn x+y=1. tính GTNN của Q= 2x^2-y^2+x+1/x+2020

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1. Q=2×2−y2+x+1x+2020Q=2×2−y2+x+1x+2020

   →Q=x2+(x−y)(x+y)+x+1x+2020→Q=x2+(x−y)(x+y)+x+1x+2020

   →Q=x2+x−y+x+1x+2020→Q=x2+x−y+x+1x+2020

   →Q=x2+x−(1−x)+x+1x+2020→Q=x2+x−(1−x)+x+1x+2020

   →Q=x2+3x+1x+2019→Q=x2+3x+1x+2019

   →Q=(x−12)2+4x+1x+2019−14→Q=(x−12)2+4x+1x+2019−14

   →Q≥0+2√4x.1x+2019−14→Q≥0+24x.1x+2019−14

   →Q≥2019+154 

    

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2. Đáp án:

    $Q\ge 2\sqrt{2020.2017}-4037$

   Giải thích các bước giải:

   $Q=\dfrac{2x^2-y^2+x+1}{x+2020}$

   $\rightarrow Q=\dfrac{x^2+(x^2-y^2)+x+1}{x+2020}$

   $\rightarrow Q=\dfrac{x^2+(x-y)(x+y)+x+1}{x+2020}$

   $\rightarrow Q=\dfrac{x^2+(x-y).1+x+1}{x+2020}$

   $\rightarrow Q=\dfrac{x^2+2x+1-y}{x+2020}$

   $\rightarrow Q=\dfrac{x^2+2x+x}{x+2020}$

   $\rightarrow Q=\dfrac{x^2+3x}{x+2020}$

   $\rightarrow Q=\dfrac{x^2-2020^2+3x+3.2020+2020^2-3.2020}{x+2020}$

   $\rightarrow Q=\dfrac{(x-2020)(x+2020)+3(x+2020)+2020(2020-3)}{x+2020}$

   $\rightarrow Q=x-2020+3+\dfrac{2020.2017}{x+2020}$

   $\rightarrow Q=x+2020+\dfrac{2020.2017}{x+2020}-4037$

   $\rightarrow Q\ge 2\sqrt{(x+2020).\dfrac{2020.2017}{x+2020}}-4037$

   $\rightarrow Q\ge 2\sqrt{2020.2017}-4037$

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