cho D=2021-1/4-2/5-3/6-......-2021/2024 và E=1/20+1/25+.....+1/10120 . so sánh D và E
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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.
24 tháng 7 2023
a: \(B=\dfrac{154}{155+156}+\dfrac{155}{155+156}\)
\(\dfrac{154}{155}>\dfrac{154}{155+156}\)
\(\dfrac{155}{156}>\dfrac{155}{155+156}\)
=>154/155+155/156>(154+155)/(155+156)
=>A>B
b: \(C=\dfrac{2021+2022+2023}{2022+2023+2024}=\dfrac{2021}{6069}+\dfrac{2022}{6069}+\dfrac{2023}{6069}\)
2021/2022>2021/6069
2022/2023>2022/2069
2023/2024>2023/6069
=>D>C
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
18 tháng 11 2023
a/
2020.2021=(2019+1)(2022-1)=
=2019.2022-2019+2022-1=2019.2022+2>2019.2022
b/
\(4^7=\left(2^2\right)^7=2^{14}< 2^{15}\)
c/
\(199^{20}< 200^{20}=\left(8.25\right)^{20}=\left(2^3.5^2\right)^{20}=2^{60}.5^{40}\)
\(2000^{15}=\left(16.125\right)^{15}=\left(2^4.5^3\right)^{15}=2^{60}.5^{45}\)
\(\Rightarrow2000^{15}=2^{60}.5^{45}>2^{60}.5^{40}>199^{20}\)
d/
\(31^{31}< 32^{31}=\left(2^5\right)^{31}=2^{155}\)
\(17^{39}>16^{39}=\left(2^4\right)^{39}=2^{156}\)
\(\Rightarrow17^{39}=2^{156}>2^{155}>31^{31}\)
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
PV
1
23 tháng 3 2023
P=[(1-2)+(-3+4)+(5-6)+(-7+8)+...+(993-994)+(-995+996)]+997
P=[(-1)+1+(-1)+1+...+(-1)+1+(-1)+1]+997
P= 0 +0 +...+ 0 +997
P=997
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
26 tháng 11 2023
a:
Sửa đề: \(S=1-3+5-7+...+2021-2023+2025\)
Từ 1 đến 2025 sẽ có:
\(\dfrac{2025-1}{2}+1=\dfrac{2024}{2}+1=1013\left(số\right)\)
Ta có: 1-3=5-7=...=2021-2023=-2
=>Sẽ có \(\dfrac{1013-1}{2}=\dfrac{1012}{2}=506\) cặp có tổng là -2 trong dãy số này
=>\(S=506\cdot\left(-2\right)+2025=2025-1012=1013\)
b: \(S=1+2-3-4+5+6-7-8+...+2021+2022-2023-2024\)
Từ 1 đến 2024 là: \(\dfrac{\left(2024-1\right)}{1}+1=2024\left(số\right)\)
Ta có: 1+2-3-4=5+6-7-8=...=2021+2022-2023-2024=-4
=>Sẽ có \(\dfrac{2024}{4}=506\) cặp có tổng là -4 trong dãy số này
=>\(S=506\cdot\left(-4\right)=-2024\)
Ta có:
\(D=2021-\dfrac{1}{4}-\dfrac{2}{5}-\dfrac{3}{6}-...-\dfrac{2021}{2024}\\ =\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{2}{5}\right)+\left(1-\dfrac{3}{6}\right)+...+\left(1-\dfrac{2021}{2024}\right)\\ =\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{3}{6}+...+\dfrac{3}{2024}\\ =3\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{2024}\right)\)
\(E=\dfrac{1}{20}+\dfrac{1}{25}+...+\dfrac{1}{10120}\\ =\dfrac{1}{4.5}+\dfrac{1}{5.5}+...+\dfrac{1}{2024.5}\\ =\dfrac{1}{5}\left(\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2024}\right)\)
Dễ thấy
\(3\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{2024}\right)>\dfrac{1}{5}\left(\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2024}\right)\\ \Rightarrow D>E\)