\(\dfrac{2021}{2022}\) x \(\dfrac{2022020222022}{202320232023}\) x \(\dfrac{20212021}{20232023}\)

= \(\dfrac{2021}{2022}\) x \(\dfrac{2022}{2023}\) x \(\dfrac{2021}{2023}\)

= \(\dfrac{2021\times2021}{2023\times2023}\)

= \(\dfrac{4084441}{4092529}\)

kết quả là 1022 nhé bạn

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

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

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

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

= 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

=2023phần2053345

\(\frac{2022\times2023-2020\times2023}{2022\times2023+2024\times7+2016}\)

\(=\frac{2023\times\left(2022-2020\right)}{2022\times2023+7\times\left(2023+1\right)+2016}\)

\(=\frac{2023\times2}{2023\times2022+7\times2023+7+2016}=\frac{2023\times2}{2023\times\left(2022+7+1\right)}=\frac{2}{2022+8}\)

\(=\frac{2}{2030}=\frac{1}{1015}\)

Với x = 2023 

<=> x + 1 = 2024

Khi đó P(2023) = x²⁰²³ - (x + 1).x²⁰²² + ... + (x + 1).x - 1

= x²⁰²³ - x²⁰²³ - x²⁰²² + .. + x² + x - 1

= x - 1 = 2023 - 1 = 2022

-2021 . 20222022 + 2022 .20212021

= -2021 . 2022. 10001 + 2022.2021. 10001

= 0

-2021 . 20222022 + 2022 . 20212021

= (-2021) . (20222022 - 2022 . 10001)

= (-2021) . (20222022 - 20222022)

= (-2021) . 0

= 0

\(V=\dfrac{1}{3}\pi r^2h=\dfrac{1}{3}\pi.2^2.3=4\pi\)

Chọn B