không làm phép tính hãy so sánh:2022^2 và 2022.2023
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`2022xx2023-2022xx1024+2022`
`=2022xx(2023-1024+1)`
`=2022xx1000`
`=2022000`
a, A = \(\dfrac{2022.2023-1}{2022.2023}\) = \(\dfrac{2022.2023}{2022.2023}\) - \(\dfrac{1}{2022.2023}\) = 1 - \(\dfrac{1}{2022.2023}\)
B = \(\dfrac{2021.2022-1}{2021.2022}\) = \(\dfrac{2021.2022}{2021.2022}\) - \(\dfrac{1}{2021.2022}\) = 1 - \(\dfrac{1}{2021.2022}\)
Vì \(\dfrac{1}{2022.2023}\) < \(\dfrac{1}{2021.2022}\)
Nên A > B
b, C = \(\dfrac{2022.2023}{2022.2023+1}\)
C = \(\dfrac{2022.2023+1-1}{2022.2023+1}\) = \(\dfrac{2022.2023+1}{2022.2023+1}\) - \(\dfrac{1}{2022.2023+1}\)
C = 1 - \(\dfrac{1}{2022.2023+1}\)
D = \(\dfrac{2023.2024}{2023.2024+1}\) = \(\dfrac{2023.2024+1-1}{2023.2024+1}\)
D = 1 - \(\dfrac{1}{2023.2024+1}\)
Vì \(\dfrac{1}{2022.2023+1}\) > \(\dfrac{1}{2023.2024+1}\)
Nên C < D
\(\dfrac{2022.2023}{2022.2023}+1=1+1=2\)
\(\dfrac{2023.2024}{2023.2024}+1=1+1=2\)
Vậy: \(\dfrac{2022.2023}{2022.2023}+1=\dfrac{2023.2024}{2023.2024}+1\)
bn tham khảo link này nhé
https://olm.vn/hoi-dap/tim-kiem?id=1300742&subject=1&q=++++++++++kh%C3%B4ng+t%C3%ADnh+k%E1%BA%BFt+qu%E1%BA%A3+c%E1%BB%A5+th%E1%BB%83+h%C3%A3y+so+s%C3%A1nh+a+v%C3%A0+b+++a=+2020+.+2020b=+2018+.+2022+++++++++
a) A= 2018 x 2022 và B= 2020 x 2020
ta có :
A=2018.2022=2020.2022-2.2022
=2020.2020+2020.2-2.2022
=2020.2020+2(2020-2022)
=2020.2020-4=B-4
=>A=B-4
hay B > A 4 đơn vị
Ta có: \(B=2020.2021.2022=\left(2021-1\right).\left(2021+1\right).2021=\left(2021-1\right)^2.2021< 2021^2.2021=A\)
`a, A = 3020 xx 3110 - 5 = 3020 xx 3109 + 3020 - 5`
`= 3020 xx 3109 + 3015 = B`.
`b, B = (2022-2)(2022+2) = 2022^2-4 < 2022^2 = A.`
a là 2020x2020 và b là 2018x2022 à bạn
nếu vậy thì b=2022x2018=(2020+2)x(2020-2)=2020x2020-4<a nha
chúc bạn học tốt
HYC-23/1/2022
Ta có: 2022^2=2022.2022
Vì 2022.2022<2022.2023
=>2022^2<2022.2023
HT
TL:
Ta có :
20222 = 2022 . 2022
Mà 2022 . 2022 < 2022 . 2023
Nên 20222 < 2022 . 2023
HT