1/2022x2023 + 1/2023x2024 + 1/2024
Những câu hỏi liên quan
1/1x2+1/2x3+1/3x4+...+1/2023x2024 ?
TRẢ LỜI NHANH . HELP ME
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{2023\times2024}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\\ =1-\dfrac{1}{2024}=\dfrac{2023}{2024}\)
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1/1*2+1/2*3+...+1/2023*2024=1-1/2+1/2-1/3+...+1/2023-1/2024
=1-1/2024=2023/2024
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\(...=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\)
\(=1-\dfrac{1}{2024}=\dfrac{2023}{2024}\)
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Xem thêm câu trả lời
B = 1/2x3 + 1/3x4 + 1/4x5 * ... x + 1/ 2022x2023
\(B=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2022.2023}\)
\(B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\)
\(B=\dfrac{1}{2}-\dfrac{1}{2023}=\dfrac{2021}{4046}\)
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s=1/1X2+1/2X3+1/3X4+1/4+.....+1/2022X2023
21 + 34 + 36 - 12
= (34 + 36) + (21 - 12)
= 70 + 9
= 79
Chúc bn học tốt!
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Tính
D = \(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+....+\dfrac{1}{2022x2023}\)
tính bằng cách thuận tiện đi mà
2022x2023-1
___________
2023x2021+2022
a)2021+2020x2022/2021x2022-1
b)2022+2021x2023/2022x2023-1
Ai giúp mình nhanh nhanh với!SOS!
a. \(\dfrac{2021+2020.2022}{2021.2022-1}\)
\(\dfrac{2021.2022-2022+2021}{2021.2022-1}=\dfrac{2021.2022-1}{2021.2022-1}=1\)
\(b.\dfrac{2022+2021.2023}{2022.2023-1}=\dfrac{2021.2023-2023+2022}{2022.2023-1}\)
\(=\dfrac{2021.2023-1}{2022.2023-1}\)
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Câu 10 (1,0 điểm)
Cho S = \(\dfrac{1}{2x3}+\dfrac{1}{4x5}+\dfrac{1}{6x7}+...+\dfrac{1}{2020x2021}+\dfrac{1}{2022x2023}\)
So sánh S với \(\dfrac{1011}{2023}\)
Tính: \(S=C^0_{2024}+\dfrac{1}{2}.C^2_{2024}+\dfrac{1}{3}.C^4_{2024}+\dfrac{1}{4}.C^6_{2024}+...+\dfrac{1}{1013}.C^{2024}_{2024}\)
Tính: \(S=C^0_{2024}+\dfrac{1}{2}.C^2_{2024}+\dfrac{1}{3}.C^4_{2024}+\dfrac{1}{4}.C^6_{2024}+...+\dfrac{1}{1013}.C^{2024}_{2024}\)