a)2022.2023-2022/2021.2022+2022
b)1999.2000-1/1998.1997+3997
c)(1-1/2).(1-1/3).(1-1/4).(1-1/5)...(1-1/2022).(1-1/2023)
help me
so sánh b=1/2022+2/2021+3/2020+...+2021/2+2022/1 VÀ c=1/2+1/3+1/4+...+1/2022+1/2023
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
so sánh 2022 mũ 2023 +1 phần 2022 mũ 2021 +1 Với 2022 mũ 2021 +1 phần 2022 mũ 2019 +1.help me, giúp với ,khó quá ko làm đc
B= 1/2 + 1/3 - 1/4 +...- 1/2022 + 1/2023 C= 1/1012 + 1/1013+...+ 1/2022 + 1/2023
Tính: B-C
mọi người sửa nhanh giúp mik vs ạ
So sánh
A = \(\dfrac{2022^{2023}+1}{2022^{2024}+1}\) và B = \(\dfrac{2022^{2022}+1}{2022^{2023}+1}\)
Trước hết ta phải chứng minh \(\dfrac{a}{b}< \dfrac{a+1}{b+1}\) (a, b ϵ N; a < b).
Thật vậy, \(\dfrac{a}{b}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{a+ab}{b^2+b}\) và \(\dfrac{a+1}{b+1}=\dfrac{\left(a+1\right)b}{\left(b+1\right)b}=\dfrac{ab+b}{b^2+b}\).
Mà theo giả thuyết là a < b nên \(\dfrac{a+ab}{b^2+b}< \dfrac{ab+b}{b^2+b}\), suy ra \(\dfrac{a}{b}< \dfrac{a+1}{b+1}\) (a, b ϵ N; a < b).
Từ đây ta có:
\(B=\dfrac{2022^{2022}+1}{2022^{2023}+1}=\dfrac{2022^{2023}+2022}{2022^{2024}+2022}=\dfrac{2022^{2023}+2021+1}{2022^{2024}+2021+1}\)
Đặt \(A_1=\dfrac{2022^{2023}+2}{2022^{2024}+2}=\dfrac{2022^{2023}+1+1}{2022^{2024}+1+1}\), rõ ràng \(A_1>A\).
Đặt \(A_2=\dfrac{2022^{2023}+3}{2022^{2024}+3}=\dfrac{2022^{2023}+2+1}{2022^{2024}+2+1}\), rõ ràng \(A_2>A_1\).
...
Đặt \(A_{2020}=\dfrac{2022^{2023}+2021}{2022^{2024}+2021}=\dfrac{2022^{2023}+2020+1}{2022^{2024}+2020+1}\), rõ ràng \(A_{2020}>A_{2019}\) và \(B>A_{2020}\).
Suy ra \(B>A_{2020}>A_{2019}>...>A_2>A_1>A\). Vậy A < B.
Ta có A = \(\dfrac{2022^{2023}}{2022^{2024}}=\dfrac{1}{2022}\) ; B = \(\dfrac{2022^{2022}}{2022^{2023}}=\dfrac{1}{2022}\)
Mà \(\dfrac{1}{2022}=\dfrac{1}{2022}\)
Vậy A = B
(1/2-1).(1/3-1).(1/4-1).(1/5-1).....(1/2022-1).(1/2023-1).
híp me!
2. Cho:
B= 1 - 1/2 + 1/3 - 1/4 +...+ 1/2021 - 1/2022 + 1/2023 C= 1/1012 + 1/1013 + 1/1014 +...+ 1/2021 + 1/2022 + 1/2023
Tính: B-C
So sánh 2 phân số
A = \(\dfrac{2022^{2022}+1}{2022^{2021}+1}\) ; B = \(\dfrac{2022^{2023}+1}{2021^{2022}+1}\)
2023-1/2*(1+2)-1/3*(1+2+3)-1/4*(1+2+3+4)-...-1/2022*(1+2+3+4+...+2022)
so sánh A = 2022^2023 + 3/2022^2022 - 1 và B = 2022^2023 - 2019/2022^2022 - 2