A = \(\dfrac{x^2-2x+2020}{2021x^2}\)

= \(\dfrac{2020x^2-2.2020.x+2020^2}{2021.2020x^2}\)

\(=\dfrac{2019x^2}{2021.2020x^2}+\dfrac{x^2-2.2020.x+2020^2}{2021.2020x^2}\)

= \(\dfrac{2019}{2021.2020}+\dfrac{\left(x-2020\right)^2}{2021.2020x^2}\ge\dfrac{2019}{2021.2020}\)

Dấu "=" xảy ra <=> x - 2020 = 0

                       <=> x = 2020

Vậy minA = \(\dfrac{2019}{2021.2020}\)đạt được tại x = 2020

\(E=\left(2x-5\right)^{10}-12\ge-12\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{5}{2}\)

Vậy \(E_{min}=-12\Leftrightarrow x=\dfrac{5}{2}\)

\(F=\left(x+5\right)^8+\left|x+5\right|+22\ge22\)

Dấu "=" xảy ra \(\Leftrightarrow x=-5\)

Vậy \(F_{min}=22\Leftrightarrow x=-5\)

\(G=17-\left|3x-2\right|\)

Dấu "=" xảy ra \(x=\dfrac{2}{3}\)

Vậy ​\(G_{max}=17\Leftrightarrow x=\dfrac{2}{3}\)​

\(K=17-\left|3x-2\right|-\left(2-3x\right)^{2020}\le17\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{2}{3}\)

Vậy \(K_{max}=17\Leftrightarrow x=\dfrac{2}{3}\)

Ta co : \(A=1-\frac{2}{x}+\frac{2020}{x^2}\)

Dat \(\frac{1}{x}=a\)ta duoc

 \(A=2020a^2-2a+1=2020\left(t-\frac{1}{2020}\right)^2+\frac{2019}{2020}\ge\frac{2019}{2020}\)

Dau "=" xay ra \(< =>x=2020\)

Vay min A = 2019/2020 khi x = 2020

Với \(x<4,\) ta có: \(A=-x+4-x+2020=2024-2x\). Do \(x<4\) nên \(A>2024-2.4=2016\).

Với \(4\le x\le2020\), ta có: \(A=x-4-x+2020=2016\).

Với \(x>2020,\) ta có \(A=x-4+x-2020=2x-2024\). Do \(x>2020\) nên \(A>2.2020-2024=2016\)

Vậy \(minA=2016\) khi \(x\in\left[4;2020\right]\)

Chúc em luôn học tập tốt :)

2016 nhé! Ủng hộ nha

\(x+y=2\Rightarrow y=2-x\)

\(P=2x^2-\left(2-x\right)^2-5x+\dfrac{1}{x}+2020=x^2-x+\dfrac{1}{x}+2016\)

\(P=x^2+1-x+\dfrac{1}{x}+2015\ge2x-x+\dfrac{1}{x}+2015\)

\(P\ge x+\dfrac{1}{x}+2015\ge2\sqrt{\dfrac{x}{x}}+2015=2017\)

Dấu "=" xảy ra khi \(x=y=1\)