So sánh:
2022.2023/ 2022.2023+1 và 2023.2024/2023.2024+1
\(\dfrac{2022.2023-1}{2022.2023}và\dfrac{2021.2022-1}{2021.2022}\)
\(\dfrac{2022.2023}{2022.2023+1}và\dfrac{2023.2024}{2023.2024+1}\)
SO SÁNH
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
so sánh 2023.2024−1/2023.2024 và 2024.2025-1/2024.2025
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
1.So sánh
a.67/77 và 73/83
b.2021.2020-1 /2021.2022 và 2022.2023-1/2022.2023
c.n+1/n+2 và n/n+3 ( n € N* )
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
Cho S= 1/2.3 + 1/4.5 + 1/6.7 +...+ 1/2020.2021 + 1/2022.2023. SO SÁNH S và 1011/2023
AI ĐÚNG MIK TICK NHƯNG PHẢI NHANH NHA!
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
không làm phép tính hãy so sánh:2022^2 và 2022.2023
Ta có: 2022^2=2022.2022
Vì 2022.2022<2022.2023
=>2022^2<2022.2023
HT
TL:
Ta có :
20222 = 2022 . 2022
Mà 2022 . 2022 < 2022 . 2023
Nên 20222 < 2022 . 2023
HT
TL:
Ta có :
20222 = 2022 . 2022
Mà 2022 . 2022 < 2022 . 2023
=> 20222 < 2022 . 2023
HT
(x+1/1.2)+(x+1/2.3)+(x+1/3.4)+....+(x+1/2022.2023)=2023x
\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)=2023x\)
\(\Rightarrow2022x+\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}+\dfrac{1}{2022}-\dfrac{1}{2023}\right)=2023x\)\(\Rightarrow2022x-2023x=-\left(1-\dfrac{1}{2023}\right)\)
\(\Rightarrow-x=-\dfrac{2022}{2023}\Leftrightarrow x=\dfrac{2022}{2023}\)
(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
Tính hợp lý : (2023.2025-976) :(2023.2024+1047)
\(\left(2023\cdot2025-976\right):\left(2023\cdot2024+1047\right)\)
\(\left(2023\cdot2024+2023-976\right):\left(2023\cdot2024+1047\right)\)
\(2023-976+1047\)
\(2023+1047-976\)
\(3070-976\)
\(2094\)
\(x-\dfrac{1}{2}-\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-...-\dfrac{1}{2021.2022}-\dfrac{1}{2022.2023}=\dfrac{-2024}{2023}\)
x-(1/1.2 + 1/2.3 + 1/3.4 + ...+ 1/2022.2023)= -2024/2023
x-(1-1/2 + 1/2-1/3 + 1/3-1/4 + ... + 1/2022-1/2023)=-2024/2023
x-(1-1/2023)=-2024/2023
x-2022/2023=-2024/2023
x = -2024/2023+2022/2023
x = -2/2023
Vậy x = -2/2023
x+2x+3x+...+2022x=2022.2023
\(x+2x+3x+...+2022x=2022.2023\\ \Leftrightarrow\left(1+2022\right).1011.x=2022.2023\\ \Leftrightarrow2023.1011.x=2022.2023\\ \Leftrightarrow1011.x=2022\\ \Leftrightarrow x=\dfrac{2022}{1011}=2\\ Vậy:x=2\)