\(A=\left|2020-x\right|+\left|2021-x\right|+\left|2022-x\right|\)

\(=\left(\left|2020-x\right|+\left|x-2022\right|\right)+\left|2021-x\right|\)

Nhận xét: \(\left\{{}\begin{matrix}\left|2020-x\right|+\left|x-2022\right|\ge\left|2020-x+x-2022\right|=2\\\left|2021-x\right|\ge0\end{matrix}\right.\)

=> \(A\ge2\)

Dấu = xảy ra khi:

\(\left\{{}\begin{matrix}\left(2020-x\right)\left(x-2022\right)\ge0\\2021-x=0\end{matrix}\right.\Leftrightarrow x=2021\)

A nhỏ nhất \(=2\Leftrightarrow x=2021\)

 a) \(\dfrac{-4}{69}\)

  b) \(\dfrac{-8}{3}\)  

câu 1: căn bậc hai của -5,5 mũ 2 bằng -5,5

câu 2:\(\left(x+7\right)^2=7\Leftrightarrow x+7=\sqrt{7}\Leftrightarrow x=\sqrt{7}-7\)

câu 3:\(\sqrt{19-x}=19\Leftrightarrow19-x=19^2\Leftrightarrow19-x=361\)\(\Leftrightarrow x=19-361=-342\)

tick cho mình nhé

2³. 19 - 2³. 14 + 1²⁰¹⁸

= 8. 19 - 8. 14 + 1

= 8. ( 19 - 14 ) + 1

= 8. 5 + 1

= 40 + 1

= 41

th1:x/2+x/3-1=2

<=>5x/6=3

<=>5x=18

<=>x=18/5

th2:x/2+x/3-1=-2

<=>5x/6=-1

<=>5x=-6

<=>x=-6/5
Vậy...

nhanh giúp mình ik mà ~ please

\(\dfrac{x}{2.4}+\dfrac{x}{4.6}+\dfrac{x}{6.8}+...+\dfrac{x}{98.100}=2\\ \Rightarrow x.\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{98.100}\right)=2\\ \Rightarrow x.\left[\dfrac{1}{2}.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{98.100}\right)\right]=2\\ \Rightarrow x.\left[\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\right]=2\Rightarrow x.\left[\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\right]=2\\ \Rightarrow x.\left[\dfrac{1}{2}.\left(\dfrac{50}{100}-\dfrac{1}{100}\right)\right]=2\\ \Rightarrow x.\left[\dfrac{1}{2}.\dfrac{49}{100}\right]=2\\ \Rightarrow x.\dfrac{49}{200}=2\\ \Rightarrow x=2:\dfrac{49}{200}\\ \Rightarrow x=2.\dfrac{200}{49}\\ \Rightarrow x=\dfrac{400}{49}\)

x/2.4+x/4.6+...+x/98.100=2

<=>x(1/2.4+1/4.6+...+1/98.100)=2

<=>1/2x(1/2-1/4+1/4-1/6+...+1/98-1/100)=2

<=>1/2x(1/2-1/100)=2

<=>1/2x49/100=2

<=>49/200x=2

<=>x=400/49