B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + \(\dfrac{2022}{1}\)

B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + 2022

B = 1 + ( 1 + \(\dfrac{1}{2022}\)) + ( 1 + \(\dfrac{2}{2021}\)) + \(\left(1+\dfrac{3}{2020}\right)\)+ ... + \(\left(1+\dfrac{2021}{2}\right)\) 

B = \(\dfrac{2023}{2023}\) + \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2021}\) + \(\dfrac{2023}{2020}\) + ...+ \(\dfrac{2023}{2}\) 

B = 2023 \(\times\) ( \(\dfrac{1}{2023}\) + \(\dfrac{1}{2022}\) + \(\dfrac{1}{2021}\) + \(\dfrac{1}{2020}\)+ ... + \(\dfrac{1}{2}\))

Vậy B > C 

A<B

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = (\(\dfrac{2021}{2}+1\))+(\(\dfrac{2020}{3}+1\))+....+(\(\dfrac{1}{2022}+1\))

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = \(\dfrac{2023}{2}\)+\(\dfrac{2023}{3}\)+....+ \(\dfrac{2023}{2022}\)

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = 2023.( \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\))

vậy x= 2023

Chọn C

C

\(\dfrac{2022}{x^2+4x+2026}=\dfrac{2022}{\left(x+2\right)^2+2022}\)

Ta có \(\left(x+2\right)^2+2022\ge2022\Leftrightarrow\dfrac{2022}{\left(x+2\right)^2+2022}\ge\dfrac{2022}{2022}=1\)

Dấu \("="\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

\(\dfrac{2022\times2023-1}{2023\times2021+2022}\)

= \(\dfrac{\left(2021+1\right)\times2023-1}{2023\times2021+2022}\)

= \(\dfrac{2023\times2021+2023-1}{2023\times2021+2022}\)

= \(\dfrac{2023\times2021+2022}{2023\times2021+2022}\)

= 1

2023×2021+20222022×2023−1​

= (2021+1)×2023−12023×2021+20222023×2021+2022(2021+1)×2023−1​

= 2023×2021+2023−12023×2021+20222023×2021+20222023×2021+2023−1​

= 2023×2021+20222023×2021+20222023×2021+20222023×2021+2022​

= 1

=>\(\left(\dfrac{2-x}{2021}-1\right)=\left(\dfrac{1-x}{2022}-1\right)+\left(1-\dfrac{x}{2023}\right)\)

=>2023-x=0

=>x=2023

giúp vơi

khó difficult