So sánh các số hữu tỉ sau:
a) 267/-268 và -1347/1343
b) 2022.2023-1/2022.2023 và 2023.2024-1/2023.2024
c) 2022.2023/2022.2023+1 và 2023.2024/2023.2024+1
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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
\(\dfrac{2022.2023}{2022.2023}+1=1+1=2\)
\(\dfrac{2023.2024}{2023.2024}+1=1+1=2\)
Vậy: \(\dfrac{2022.2023}{2022.2023}+1=\dfrac{2023.2024}{2023.2024}+1\)
Ta có:2023.2024<2024.2025
=> 1/2023.2024>1/2024.2025
=>-1/2023.2024<-1/2024.2025
=> 2023.2024-1/2023.2024<2024.2025-1/2024.2025
Học tốt
a. 67/77 = 1 - 10/77; 73/83=1 - 10/83
Vì 10/77>10/83 nên 1 - 10/77 < 1-10/83
Vậy 67/77<73/83
c. Ta có: n/n+3 < n+1/n+3 <n+1/n+2
Vậy n/n+3 < n+1/n+2
-5<0<1/63
-101/-100=101/100>1>200/201
1/17>1/27>3/83
135/136=1+(1/136)>1+(1/137)=136/137
-371/459<0<-371/-459
267/-268>-1>-1347/1343
-13/38<-1/3<29/-88
-18/31=\(\frac{-18.10101}{31.10101}=\frac{-181818}{313131}\)
Yêu cầu đề là gì vậy bạn? Bạn nên ghi rõ ràng, đầy đủ để mọi người hỗ trợ tốt hơn/
\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)=2023x\)
\(\Rightarrow2022x+\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}+\dfrac{1}{2022}-\dfrac{1}{2023}\right)=2023x\)\(\Rightarrow2022x-2023x=-\left(1-\dfrac{1}{2023}\right)\)
\(\Rightarrow-x=-\dfrac{2022}{2023}\Leftrightarrow x=\dfrac{2022}{2023}\)
(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
S=1/2x3+1/4x5+1/6x7+...+1/2022x2023<1/2x3+1/3x4+1/4x5+...+1/1010x1011
=1/2-1/1011=1009/2022<1011/2023
=>S<1011/2023
S= 1/2.3 + 1/4.5 + 1/6.7 +.....+ 1 2020.2021 + 1 2022.2023 . : So sánh S và 1011/2023
tui làm được câu c thui
c) (1-1/2).(1-1/3).(1-1/4).(1-1/5)...(1-1/2022).(1-1/2023)
a)
\(\dfrac{267}{268}< 1\Rightarrow-\dfrac{267}{268}>-1\)
\(\dfrac{1347}{1343}>1\Rightarrow-\dfrac{1347}{1343}< -1\)
\(\Rightarrow-\dfrac{1347}{1343}< -\dfrac{267}{268}\)
b) \(\dfrac{2022\cdot2023-1}{2022\cdot2023}=\dfrac{2022\cdot2023}{2022\cdot2023}-\dfrac{1}{2022\cdot2023}=1-\dfrac{1}{2022\cdot2023}\)
\(\dfrac{2023\cdot2024-1}{2023\cdot2024}=\dfrac{2023\cdot2024}{2023\cdot2024}-\dfrac{1}{2023\cdot2024}=1-\dfrac{1}{2023\cdot2024}\)
Vì: \(2022\cdot2023< 2023\cdot2024\)
\(\Rightarrow\dfrac{1}{2022\cdot2023}>\dfrac{1}{2023\cdot2024}\)
\(\Rightarrow1-\dfrac{1}{2022\cdot2023}< 1-\dfrac{1}{2023\cdot2024}\)
Hay: `(2022*2023-1)/(2022*2023) < (2023*2024 - 1)/(2023*2024)`
c) \(\dfrac{2022\cdot2023}{2022\cdot2023+1}=\dfrac{2023\cdot2023+1-1}{2022\cdot2023+1}=1-\dfrac{1}{2022\cdot2023+1}\)
\(\dfrac{2023\cdot2024}{2023\cdot2024+1}=\dfrac{2023\cdot2024+1-1}{2023\cdot2024+1}=1-\dfrac{1}{2023\cdot2024+1}\)
Vì: \(2022\cdot2023+1< 2023\cdot2024+1\)
\(\Rightarrow\dfrac{1}{2022\cdot2023+1}>\dfrac{1}{2023\cdot2024+1}\)
\(\Rightarrow1-\dfrac{1}{2022\cdot2023+1}< 1-\dfrac{1}{2023\cdot2024+1}\)
Hay: `(2022*2023)/(2022*2023+1)<(2023*2024)/(2023*2024+1)`