Toán cho 2 số thực x, y dương thoả mãn x+y=1. tính GTNN của Q= 2x^2-y^2+x+1/x+2020 03/08/2021 By Alaia cho 2 số thực x, y dương thoả mãn x+y=1. tính GTNN của Q= 2x^2-y^2+x+1/x+2020
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 Trả lời
Đáp án: $Q\ge 2\sqrt{2020.2017}-4037$ Giải thích các bước giải: $Q=\dfrac{2x^2-y^2+x+1}{x+2020}$ $\rightarrow Q=\dfrac{x^2+(x^2-y^2)+x+1}{x+2020}$ $\rightarrow Q=\dfrac{x^2+(x-y)(x+y)+x+1}{x+2020}$ $\rightarrow Q=\dfrac{x^2+(x-y).1+x+1}{x+2020}$ $\rightarrow Q=\dfrac{x^2+2x+1-y}{x+2020}$ $\rightarrow Q=\dfrac{x^2+2x+x}{x+2020}$ $\rightarrow Q=\dfrac{x^2+3x}{x+2020}$ $\rightarrow Q=\dfrac{x^2-2020^2+3x+3.2020+2020^2-3.2020}{x+2020}$ $\rightarrow Q=\dfrac{(x-2020)(x+2020)+3(x+2020)+2020(2020-3)}{x+2020}$ $\rightarrow Q=x-2020+3+\dfrac{2020.2017}{x+2020}$ $\rightarrow Q=x+2020+\dfrac{2020.2017}{x+2020}-4037$ $\rightarrow Q\ge 2\sqrt{(x+2020).\dfrac{2020.2017}{x+2020}}-4037$ $\rightarrow Q\ge 2\sqrt{2020.2017}-4037$ Trả lời
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
Đáp án:
$Q\ge 2\sqrt{2020.2017}-4037$
Giải thích các bước giải:
$Q=\dfrac{2x^2-y^2+x+1}{x+2020}$
$\rightarrow Q=\dfrac{x^2+(x^2-y^2)+x+1}{x+2020}$
$\rightarrow Q=\dfrac{x^2+(x-y)(x+y)+x+1}{x+2020}$
$\rightarrow Q=\dfrac{x^2+(x-y).1+x+1}{x+2020}$
$\rightarrow Q=\dfrac{x^2+2x+1-y}{x+2020}$
$\rightarrow Q=\dfrac{x^2+2x+x}{x+2020}$
$\rightarrow Q=\dfrac{x^2+3x}{x+2020}$
$\rightarrow Q=\dfrac{x^2-2020^2+3x+3.2020+2020^2-3.2020}{x+2020}$
$\rightarrow Q=\dfrac{(x-2020)(x+2020)+3(x+2020)+2020(2020-3)}{x+2020}$
$\rightarrow Q=x-2020+3+\dfrac{2020.2017}{x+2020}$
$\rightarrow Q=x+2020+\dfrac{2020.2017}{x+2020}-4037$
$\rightarrow Q\ge 2\sqrt{(x+2020).\dfrac{2020.2017}{x+2020}}-4037$
$\rightarrow Q\ge 2\sqrt{2020.2017}-4037$