So sánh cặp số sau
a) \(\dfrac{-22}{45}\) và \(\dfrac{-51}{103}\) b)\(\dfrac{2020^{2014}+1}{2020^{2015}+1}\) và \(\dfrac{2020^{2015}-2}{2020^{2016}-2}\)
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11 tháng 5 2023
\(A>\dfrac{2^{2018}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{3^{2019}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{5^{2020}}{5^{2020}+2^{2018}+3^{2019}}=1\)
\(B< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2019\cdot2020}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\)
=>B<1
=>A>B
BD
16 tháng 7 2023
a) Ta có:
2A=2.(12+122+123+...+122020+122021)2�=2.12+122+123+...+122 020+122 021
2A=1+12+122+123+...+122019+1220202�=1+12+122+123+...+122 019+122 020
Suy ra: 2A−A=(1+12+122+123+...+122019+122020)2�−�=1+12+122+123+...+122 019+122 020
−(12+122+123+...+122020+122021)−12+122+123+...+122 020+122 021
Do đó A=1−122021<1�=1−122021<1.
Lại có B=13+14+15+1360=20+15+12+1360=6060=1�=13+14+15+1360=20+15+12+1360=6060=1.
Vậy A < B.
8 tháng 3 2021
Bài 2:
Ta có: \(11^{1979}< 11^{1980}=1331^{660}\)
\(37^{1320}=37^{2\cdot660}=1369^{660}\)
mà \(1331^{660}< 1369^{660}\)
nên \(11^{1979}< 37^{1320}\)
17 tháng 4 2021
Giải:
Ta có: N=2019+2020/2020+2021
=>N=2019/2020+2021 + 2020/2020+2021
Vì 2019/2020 > 2019/2020+2021 ; 2020/2021 > 2020/2020+2021
=>M>N
Vậy ...
Chúc bạn học tốt!
17 tháng 4 2021
Ta có : \(\dfrac{2019}{2020}>\dfrac{2019}{2020+2021}\)
\(\dfrac{2020}{2021}>\dfrac{2020}{2020+2021}\)
\(\Rightarrow\dfrac{2019}{2020}+\dfrac{2020}{2021}>\dfrac{2019+2020}{2020+2021}\)
\(\Rightarrow M>N\)
15 tháng 8 2023
c: \(100C=\dfrac{100^{100}+100}{100^{100}+1}=1+\dfrac{99}{100^{100}+1}\)
\(100D=\dfrac{100^{101}+100}{100^{101}+1}=1+\dfrac{99}{100^{101}+1}\)
100^100+1<100^101+1
=>\(\dfrac{99}{100^{100}+1}>\dfrac{99}{100^{101}+1}\)
=>100C>100D
=>C>D
b: \(2020E=\dfrac{2020^{2022}+2020}{2020^{2022}+1}=1+\dfrac{2019}{2020^{2022}+1}\)
\(2020F=\dfrac{2020^{2021}+2020}{2020^{2021}+1}=1+\dfrac{2019}{2020^{2021}+1}\)
2020^2022+1>2020^2021+1(Do 2022>2021)
=>\(\dfrac{2019}{2020^{2022}+1}< \dfrac{2019}{2020^{2021}+1}\)
=>2020E<2020F
=>E<F
25 tháng 2 2022
\(2.A=\frac{2^{2021}-2}{2^{2021}-1}=1-\frac{1}{2^{2021}-1}\)
\(2B=\frac{2^{2022}-2}{2^{2022}-1}=1-\frac{1}{2^{2022}-1}\)
dó \(\frac{1}{2^{2022}-1}< \frac{1}{2^{2021}-1}\Rightarrow1-\frac{1}{2^{2022}-1}>1-\frac{1}{2^{2021}-1}\Rightarrow A< B\)
HT
a: \(\dfrac{-22}{45}=\dfrac{-22\cdot103}{45\cdot103}=\dfrac{-2266}{45\cdot103}\)
\(\dfrac{-51}{103}=\dfrac{-51\cdot45}{45\cdot103}=-\dfrac{2295}{45\cdot103}\)
mà -2266>-2295
nên \(-\dfrac{22}{45}>-\dfrac{51}{103}\)
b: Đặt \(A=\dfrac{2020^{2014}+1}{2020^{2015}+1};B=\dfrac{2020^{2015}-2}{2020^{2016}-2}\)
\(2020A=\dfrac{2020^{2015}+2020}{2020^{2015}+1}=1+\dfrac{2019}{2020^{2015}+1}\)
\(2020B=\dfrac{2020^{2016}-4040}{2020^{2016}-2}=1-\dfrac{4038}{2020^{2016}-2}\)
mà \(\dfrac{2019}{2020^{2015}+1}>\dfrac{-4038}{2020^{2016}-2}\)
nên 2020A>2020B
=>A>B