Cho minh hoi bai nay vs:
Chung to S=\(\dfrac{1}{4}\)+\(\dfrac{2}{4}\)+...+\(\dfrac{10}{4}\)<14
Chung to S=\(\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{19}< 2\)
Chung to S=\(\dfrac{1}{10}+\dfrac{1}{4}+...+\dfrac{1}{100}>1\)
K
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DK
18 tháng 9 2022
Ta có:
1/2^2 > 1/2.3
1/3^2 > 1/3.4
...
1/10^2 > 1/10.11
-> Cộng dọc theo vế ta có:
1/2^2+1/3^2+...+1/10^2 > 1/2.3+1/3.4+...+1/10.11
= 1/2-1/3+1/3-1/4+...+1/10-1/11
= 1/2 - 1/11 = 9/22 (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 < \(\dfrac{1}{2}\)
24 tháng 4 2022
4S=1+24+342+....+2014420134S=1+24+342+....+201442013
4S−S=3S=1+24+342+....+201442013−(14+242+343+....+201442014)4S−S=3S=1+24+342+....+201442013−(14+242+343+....+201442014)
3S=1+(24−14)+(342−242)+......+(201442013−201342013)−2014420143S=1+(24−14)+(342−242)+......+(201442013−201342013)−201442014
3S=1+14+142+143+.....+142013−2014420143S=1+14+142+143+.....+142013−201442014
đặt A=1+14+142+143+....+142023A=1+14+142+143+....+142023
4A−A=4+1+14+142+.....+142022−(1+14+142+....+142023)4A−A=4+1+14+142+.....+142022−(1+14+142+....+142023)
3A=4−1420233A=4−142023
A=43−13.42023A=43−13.42023
⇒3S=43−13.42023−201442024⇒3S=43−13.42023−201442024
⇒S=49−19.42023−20143.42024⇒S=49−19.42023−20143.42024
do 49<48=1249<48=12
⇒S=49−19.42023−20143.42024<48=12(đpcm)
29 tháng 6 2022
Bài 1:
a: \(P=\left(\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2}{\left(x+1\right)^2}\right)\cdot\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{4}\)
\(=\dfrac{x^2-x-2-x^2-x+2}{\left(x-1\right)\left(x+1\right)^2}\cdot\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{4}\)
\(=\dfrac{-2x}{1}\cdot\dfrac{x-1}{4}=-\dfrac{x\left(x-1\right)}{2}\)
b: Để \(\dfrac{P-4}{5}=x\) thì P-4=5x
=>P=5x+4
\(\Leftrightarrow-\dfrac{x\left(x-1\right)}{2}=5x+4\)
=>-x2+x=10x+8
=>x2-x=-10x-8
=>x2+9x+8=0
=>x=-8(nhận) hoặc x=-1(loại)
13 tháng 5 2017
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{200^2}>\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{199.200}=\dfrac{199}{200}>\dfrac{3}{4}\)
LT
Lương Thị Vân Anh
CTVHS
16 tháng 4 2023
Ta có S = \(\dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3}{8}+\dfrac{4}{16}+...+\dfrac{10}{2^{10}}\)
= \(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{10}{2^{10}}\)
2S = 1 + \(\dfrac{2}{2}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{10}{2^9}\)
2S - S = ( 1 + \(\dfrac{2}{2}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{10}{2^9}\)) - ( \(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{10}{2^{10}}\))
S = 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}-\dfrac{10}{2^{10}}\)
Đặt A = 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\)
2A = 2 + 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^8}\)
2A - A = ( 2 + 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^8}\)) - ( 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\))
A = 2 - \(\dfrac{1}{2^9}\)
⇒ S = 2 - \(\dfrac{1}{2^9}\) - \(\dfrac{10}{2^{10}}\) = \(\dfrac{2^{11}}{2^{10}}-\dfrac{2}{2^{10}}-\dfrac{10}{2^{10}}=\dfrac{2^2\left(2^9-3\right)}{2^{10}}=\dfrac{2^9-3}{2^8}\)
Vậy S = \(\dfrac{2^9-3}{2^8}\)
— S = 1/4 + 2/4 +...+10/4 (1)
= 1 + 1/4 + 2/4 +...+ 9/4 (2)
=> Lấy (2) trừ đi (1) ta được:
1 — 10/4 = —6/4
Vì 14 = 14/1 = 84/6 mà —6/4 < 84/6
Do đó S < 14
Cậu có có thể giúp mk 2 câu tiếp theo đc ko