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\(\dfrac{2022.2023}{2022.2023}+1=1+1=2\)
\(\dfrac{2023.2024}{2023.2024}+1=1+1=2\)
Vậy: \(\dfrac{2022.2023}{2022.2023}+1=\dfrac{2023.2024}{2023.2024}+1\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)=2023x\)
\(\Rightarrow2022x+\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}+\dfrac{1}{2022}-\dfrac{1}{2023}\right)=2023x\)\(\Rightarrow2022x-2023x=-\left(1-\dfrac{1}{2023}\right)\)
\(\Rightarrow-x=-\dfrac{2022}{2023}\Leftrightarrow x=\dfrac{2022}{2023}\)
(x + 1/1.2) + (x + 1/2.3) + (x + 1/3.4) + ... + (x + 1/2022.2023) = 2023x
x + x + x + ... + x + 1/1.2 + 1/2.3 + ... + 1/2022.2023 = 2023x
2022x + 1 - 1/2 + 1/2 - 1/3 + ... + 1/2022 - 2023 = 2023x
2023x - 2022x = 1 - 1/2023
x = 2022/2023
a) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\)
\(2A=2\cdot\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{101}}\)
\(2A-A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{100}}\)
\(A=1-\dfrac{1}{2^{100}}\)
b) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2023\cdot2024}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\)
\(=1-\dfrac{1}{2024}\)
\(=\dfrac{2024}{2024}-\dfrac{1}{2024}\)
\(=\dfrac{2023}{2024}\)
Lời giải:
$A=1+2.3+3.4+4.5+...+2022.2023$
$3A=3+2.3(4-1)+3.4(5-2)+4.5(6-3)+....+2022.2023(2024-2021)$
$=3+2.3.4+3.4.5+4.5.6+...+2022.2023.2024-(1.2.3+2.3.4+3.4.5+...+2021.2022.2023)$
$=3+2022.2023.2024-1.2.3=2022.2023.2024-3$
$\Rightarrow A=2759728047$
So sánh A=\(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{2021}\)và B=20. So sánh A và B
Bài 1:
a: Sửa đề: 1/3^200
1/2^300=(1/8)^100
1/3^200=(1/9)^100
mà 1/8>1/9
nên 1/2^300>1/3^200
b: 1/5^199>1/5^200=1/25^100
1/3^300=1/27^100
mà 25^100<27^100
nên 1/5^199>1/3^300
\(S=\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}\)
Ta có :
+) \(\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}\)
+) \(\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{40}+\dfrac{1}{40}\)
\(\Leftrightarrow S< \dfrac{1}{5}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{40}+\dfrac{1}{40}\)
\(\Leftrightarrow S< \dfrac{1}{2}\)
Vậy,,,
Ta có: \(\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}=\dfrac{2}{8}=\dfrac{1}{4}\)
\(\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{40}+\dfrac{1}{40}=\dfrac{2}{40}=\dfrac{1}{20}\)
Do đó: \(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{4}+\dfrac{1}{20}=\dfrac{6}{20}=\dfrac{3}{10}\)
\(\Leftrightarrow\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{3}{10}+\dfrac{1}{5}=\dfrac{3}{10}+\dfrac{2}{10}=\dfrac{1}{2}\)
hay \(S< \dfrac{1}{2}\)(đpcm)
Bạn thiếu đề rồi phải là trừ hay cộng j j chứ.
Xét:
`A+B=2+1/2+1/3+1/4+......+1/4026+1/3+1/5+1/7+......+1/4025`
`1/2+1/3+1/4+......+1/4026+1/3+1/5+1/7+......+1/4025>0`
`=>A+B>2`
Mà `1 2013/2014<2`
`=>A+B>1 2013/2014`
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.
a, A = \(\dfrac{2022.2023-1}{2022.2023}\) = \(\dfrac{2022.2023}{2022.2023}\) - \(\dfrac{1}{2022.2023}\) = 1 - \(\dfrac{1}{2022.2023}\)
B = \(\dfrac{2021.2022-1}{2021.2022}\) = \(\dfrac{2021.2022}{2021.2022}\) - \(\dfrac{1}{2021.2022}\) = 1 - \(\dfrac{1}{2021.2022}\)
Vì \(\dfrac{1}{2022.2023}\) < \(\dfrac{1}{2021.2022}\)
Nên A > B
b, C = \(\dfrac{2022.2023}{2022.2023+1}\)
C = \(\dfrac{2022.2023+1-1}{2022.2023+1}\) = \(\dfrac{2022.2023+1}{2022.2023+1}\) - \(\dfrac{1}{2022.2023+1}\)
C = 1 - \(\dfrac{1}{2022.2023+1}\)
D = \(\dfrac{2023.2024}{2023.2024+1}\) = \(\dfrac{2023.2024+1-1}{2023.2024+1}\)
D = 1 - \(\dfrac{1}{2023.2024+1}\)
Vì \(\dfrac{1}{2022.2023+1}\) > \(\dfrac{1}{2023.2024+1}\)
Nên C < D