bạn cho đầy đủ câu hỏi đi

A=1−2−3+4−5−6+7−8−9+....+2020−2021−2022D=1-2-3+4-5-6+7-8-9+....+2020-2021-2022

A =(1−2−3)+(4−5−6)+(7−8−9)+....+(2020−2021−2022)D=(1-2-3)+(4-5-6)+(7-8-9)+....+(2020-2021-2022)

A=(−4)+(−7)+(−10)+.....+(−2023)D=(-4)+(-7)+(-10)+.....+(-2023)

A=[(2023−4):3+1].[(−2023−4):2]D=[(2023-4):3+1].[(-2023-4):2]

A=674.(−1013,5)D=674.(-1013,5)

A=−683099

A=1−2−3+4−5−6+7−8−9+....+2020−2021−2022D=1-2-3+4-5-6+7-8-9+....+2020-2021-2022

A =(1−2−3)+(4−5−6)+(7−8−9)+....+(2020−2021−2022)D=(1-2-3)+(4-5-6)+(7-8-9)+....+(2020-2021-2022)

A=(−4)+(−7)+(−10)+.....+(−2023)D=(-4)+(-7)+(-10)+.....+(-2023)

A=[(2023−4):3+1].[(−2023−4):2]D=[(2023-4):3+1].[(-2023-4):2]

A=674.(−1013,5)D=674.(-1013,5)

A=−683099

`2022+2022:1/3+2022+2022:1/5`

`=2022+2022xx3+2022+2022xx5`

`=2022xx(1+3+1+5)`

`=2022xx10=20220`

20220

a: \(\dfrac{4}{9}=\dfrac{4\cdot2}{9\cdot2}=\dfrac{8}{18}< \dfrac{13}{18}\)

b: 34/-4=-8,5

Ta có: 8,5<8,6

=>-8,5>-8,6

=>\(\dfrac{34}{-4}>-8,6\)

c: \(\dfrac{2021}{2022}=1-\dfrac{1}{2022}\)

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

Ta có: 2022<2023

=>\(\dfrac{1}{2022}>\dfrac{1}{2023}\)

=>\(-\dfrac{1}{2022}< -\dfrac{1}{2023}\)

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

=>\(\dfrac{2021}{2022}< \dfrac{2022}{2023}\)

34/-4=-8,5 là sao v

chờ tí

đâu đúng mà 

\(=\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{2}-\dfrac{5}{6}+2022=2022\)

2022

A = ( 2022 x 0,75 + 2022 : 4) - ( 2021 : 0,1 : 10)

= (2022 x 0,75 + 2022 x 0,25) - ( 2021 : 1/10 : 10)

= 2022 x(0,75+0,25) - 2021x 10 : 10

= 2022x 1 - 2021

= 2022 - 2021

=1

a) 17.13+17.42-17.35

=17.(13+42-35)

=17.20=340

b) [25.(18-4²)-10]:4+6

=(25.2-10):4+6

=40:4+6=16

c) 3⁶:3²+2³.2²-3².3

=3⁴+2⁵-3³

=81+32-27=86

d) B=3.4²-2².3

=3.(16-4)

=3.12=36

e)2022⁰+3.[5².10-(23-13)²]

=1+3.(250-100)

=1+450=451

g) 27.77+24.27-27

=27.(77+24-1)

=27.100=2700

h) 5.2³+7⁹:7⁷-1²⁰²⁰

=40+7²-1

=89-1=88

i) 120:{54[50:2+(3²-2.4)]}

=120:[54(25+1)]

=120:1404=10/117