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PH
2
H
15 tháng 3 2023
`3/4-(2/3+3/4)+2/3+2022/2023`
`=3/4 - 2/3 - 3/4 +2/3 +2022/2023`
`= (3/4 -3/4 ) + (-2/3 +2/3) +2022/2023`
`= 0+0+2022/2023`
`=2022/2023`
NK
15 tháng 3 2023
\(\dfrac{3}{4}-\left(\dfrac{2}{3}+\dfrac{3}{4}\right)+\dfrac{2}{3}+\dfrac{2022}{2023}\)
\(=\dfrac{3}{4}-\left(\dfrac{8}{12}+\dfrac{9}{12}\right)+\dfrac{2}{3}+\dfrac{2022}{2023}\)
\(=\dfrac{3}{4}-\dfrac{17}{12}+\dfrac{2}{3}+\dfrac{2022}{2023}\)
\(=\dfrac{9}{12}-\dfrac{17}{12}+\dfrac{8}{12}+\dfrac{2022}{2023}\)
\(=\dfrac{9-17+8}{12}+\dfrac{2022}{2023}=\dfrac{0}{12}+\dfrac{2022}{2023}=0+\dfrac{2022}{2023}\)
\(=\dfrac{2022}{2023}\)
#YTVA
5 tháng 6 2023
\(A=\dfrac{2}{3}+\dfrac{2}{3^2}+\dfrac{2}{3^3}+....+\dfrac{2}{3^{2023}}\)
\(3A=2+\dfrac{2}{3}+\dfrac{2}{3^2}+....+\dfrac{2}{3^{2022}}\)
\(3A-A=\left(2+\dfrac{2}{3}+\dfrac{2}{3^2}+...+\dfrac{2}{3^{2022}}\right)-\left(\dfrac{2}{3}+\dfrac{2}{3^2}+....+\dfrac{2}{3^{2023}}\right)\)
\(2A=2-\dfrac{2}{3^{2023}}\)
\(A=\left(2-\dfrac{2}{3^{2023}}\right)\times\dfrac{1}{2}\)
\(A=2\times\dfrac{1}{2}-\dfrac{2}{3^{2023}}\times\dfrac{1}{2}\)
\(A=1-\dfrac{1}{3^{2023}}\)
=> \(A< 1\left(đpcm\right)\)
KV
20 tháng 12 2023
Đặt B = 2² + 2³ + 2⁴ + ... + 2²⁰²³
⇒ 2B = 2³ + 2⁴ + 2⁵ + ... + 2²⁰²⁴
⇒ B = 2B - B
= (2³ + 2⁴ + 2⁵ + ... + 2²⁰²⁴) - (2² + 2³ + 2⁴ + ... + 2²⁰²³)
= 2²⁰²⁴ - 2²
⇒ A = 2² + 2²⁰²⁴ - 2² = 2²⁰²⁴
= 2.2²⁰²³ ⋮ 2²⁰²³
Vậy A ⋮ 2²⁰²³
AH
Akai Haruma
Giáo viên
20 tháng 12 2023
Lời giải:
$A=4+2^2+2^3+....+2^{2023}$
$2A=8+2^3+2^4+...+2^{2024}$
$\Rightarrow 2A-A=(8+2^3+2^4+...+2^{2024})-(4+2^2+2^3+....+2^{2023})$
$\Rightarrow A=2^{2024}+8-4-2^2=2^{2024}\vdots 2^{2023}$
Ta có đpcm/
VD
16 tháng 4 2023
=> 4S = 1 + 2/4 + 3/4^2 +...+ 2023/4^2022
=> 4S-S = 1 + (2/4-1/4) + (3/4^2 - 2/4^2) +...+ (2023/4^2022 - 2022/4^2022) - 2023/4^2023
=> 3S = 1 + 1/4 + 1/4^2 +...+ 1/4^2022 - 2023/4^2023
=> 12S = 4 + 1 + 1/4 +... + 1/4^2021 - 2023/4^2022
=> 12S - 3S = 4 + (1-1) + (1/4-1/4) +... + (1/4^2021 - 1/4^2021) - 1/4^2022 - 2023/4^2022 + 2023/4^2023
=> 9S = 4 - 1/4^2022 - 2023/4^2022 + 2023/4^2023
= 4- 2024/4^2022 + 2023/4^2023
Do 2024/4^2022 > 2024/4^2023 > 2023/4^2023 nên - 2024/4^2022 + 2023/4^2023 < 0
=> 9S < 4 < 9/2
=> S < 1/2 (đpcm)
LT
Lương Thị Vân Anh
CTVHS
16 tháng 4 2023
Ta có S = \(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2023}{4^{2023}}\)
4S = \(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2023}{4^{2022}}\)
4S - S = ( \(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2023}{4^{2022}}\) ) - ( \(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2023}{4^{2023}}\))
3S = 1 + \(\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2022}}-\dfrac{2023}{4^{2023}}\)
Đặt A = 1 + \(\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2022}}\)
4A = 4 + 1 + \(\dfrac{1}{4}+...+\dfrac{1}{4^{2021}}\)
4A - A = ( 4 + 1 + \(\dfrac{1}{4}+...+\dfrac{1}{4^{2021}}\)) - ( 1 + \(\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2022}}\))
3A = 4 - \(\dfrac{1}{4^{2022}}\)
A = ( 4 - \(\dfrac{1}{4^{2022}}\)) : 3 = \(\dfrac{4}{3}-\dfrac{1}{4^{2022}\cdot3}\)
⇒ 3S = \(\dfrac{4}{3}-\dfrac{1}{4^{2022}\cdot3}\) - \(\dfrac{2023}{4^{2023}}\)
S = ( \(\dfrac{4}{3}-\dfrac{1}{4^{2022}\cdot3}\) - \(\dfrac{2023}{4^{2023}}\)) : 3 = \(\dfrac{4}{9}-\dfrac{1}{4^{2022}\cdot3^2}-\dfrac{1}{4^{2023}\cdot3}< \dfrac{4}{9}< \dfrac{1}{2}\)
Vậy S < \(\dfrac{1}{2}\)
L
3
HM
2
8 tháng 3
M=(1/5+1/5^2+1/5^3+...+1/5^2023) + 1/5x(1/5+1/5^2+1/5^3+...+1/5^2022) + ... + 1/5^2021x(1/5+1/5^2) + 1/5^2022x1/5
Xét biểu thức N=1/5+1/5^2+1/5^3 + ... + 1/5^k (K>0, k thuộc Z)
=> 5N=1+1/5+1/5^2+1/5^3+...+1/5^(k-1)
=> 4N= 5N - N =1 - 1/5^k
=> 1/5+1/5^2+1/5^3 + ... + 1/5^k = 1/4x(1-1/5^k)
Thay vào biểu thức M, ta có:
M= 1/4x(1-1/5^2023) + 1/5x1/4x(1-1/5^2022) + ... + 1/5^2021x1/4x(1-1/5^2) + 1/5^2022x1/4x(1-1/5)
=> 4M = (1+1/5+1/5^2+...+1/5^2022) - 2023/5^2023
=> 4M = 5/4x(1-1/5^2023)-2023/5^2023 < 5/4
=> M < 5/16 < 1/3
Vậy M < 1/3 [ vượt chỉ tiêu nhé =)) ]
\(P=3+2^2+2^3+...+2^{2023}\)
\(2P=2\cdot\left(3+2^2+2^3+...+2^{2023}\right)\)
\(2P=6+2^3+2^4+...+2^{2024}\)
\(2P-P=\left(6+2^3+2^4+...+2^{2024}\right)-\left(3+2^2+2^3+...+2^{2023}\right)\)
\(P=\left(2^3-2^3\right)+\left(2^4-2^4\right)+...+\left(6-3\right)+\left(2^{2024}-2^2\right)\)
\(P=3+2^{2024}-2^2\)
\(P=3+2^{2024}-4\)
\(P=2^{2024}-1\)
thanks bạn rất nhiều