C=(1/2-1)(1/3-1)(1/4-1)...(1/2023-1)
K
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Những câu hỏi liên quan
11 tháng 4 2023
1+1/2.(1+2)+1/3.(1+2+3)+1/4.(1+2+3+4)+...+1/2023.(1+2+3+...+2023)
=1+1/2.(1+2).2/2+1/3.(1+3).3/2+1/4.(1+4).4/2+...+1/2023.(1+2+3+...+2023).2023/2
=2/2+3/2+4/2+...+2023/2
=2+3+4+...+2023/2
=2025.2022/2/2
=1023637,5
tham khảo thôi nha
DT
Dang Tung
CTVHS
19 tháng 8 2023
\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2023}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2022}{2023}\\ =\dfrac{1}{2023}\)
TN
1
D
24 tháng 9
Ta có: C = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/2021.2022.2023
=> C = 1/2. (3-1/1.2.3 + 4-2/2.3.4 + 5-3/3.4.5 + ... + 2023-2021/2021.2022.2023
=> C = 1/2. (1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 + ... + 1/2021.2022 - 1/2022.2023)
=> C = 1/2. (1/1.2 - 1/2022.2023)
- Phần còn lại bạn tự tính chứ số to quá
17 tháng 8 2023
1:
a: =23/27-11/17+4/27+28/17
=23/27+4/27+28/17-11/17
=1+1=2
b: \(=\dfrac{2}{3}\cdot\left(\dfrac{7}{9}+\dfrac{2}{9}\right)-\dfrac{2}{9}\)
=2/3-2/9
=6/9-2/9
=4/9
c: \(=\dfrac{11}{5}\cdot\dfrac{7}{3}-\dfrac{1}{3}\cdot\dfrac{11}{5}\)
=11/5(7/3-1/3)
=11/5*2
=22/5
d: \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2024}{2023}=\dfrac{2024}{2}=1012\)
e: \(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2022}{2023}=\dfrac{1}{2023}\)
BB
2
AH
Akai Haruma
Giáo viên
31 tháng 1
Lời giải:
\(C=(\frac{1}{2^2}-1)(\frac{1}{3^2}-1)(\frac{1}{4^2}-1)....(\frac{1}{2023^2}-1)\)
\(=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.\frac{1-4^2}{4^2}....\frac{1-2023^2}{2023^2}\)
\(=\frac{(2^2-1)(3^2-1)(4^2-1)....(2023^2-1)}{2^2.3^2.4^2....2023^2}\)
\(=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)....(2023-1)(2023+1)}{2^2.3^2.4^2....2023^2}\)
\(=\frac{1.3.2.4.3.5.....2022.2024}{(2.3.4...2023)(2.3.4...2023)}\)
\(=\frac{(1.2.3...2022)(3.4.5....2024)}{(2.3...2023)(2.3.4...2023)}\)
\(=\frac{1}{2023}.\frac{2024}{2}=\frac{1012}{2023}\)
MM
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
3 tháng 5 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
NH
0
ai biết giúp mình với nha !
`Answer:`
\(C=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2023}-1\right)\)
\(=\left(\frac{1}{2}-\frac{2}{2}\right)\left(\frac{1}{3}-\frac{3}{3}\right)\left(\frac{1}{4}-\frac{4}{4}\right)...\left(\frac{1}{2023}-\frac{2023}{2023}\right)\)
\(=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)\left(-\frac{3}{4}\right)...\left(-\frac{2022}{2023}\right)\)
\(=\frac{1.2.3...2022}{2.3.4...2023}\)
\(=\frac{1}{2023}\)