cứu tui

\(\dfrac{2022.2023}{2022.2023}+1=1+1=2\)

\(\dfrac{2023.2024}{2023.2024}+1=1+1=2\)

Vậy: \(\dfrac{2022.2023}{2022.2023}+1=\dfrac{2023.2024}{2023.2024}+1\)

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

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

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)

Lời giải:

$S=5(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2022.2023})$

$=5(\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2023-2022}{2022.2023})$

$=5(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2022}-\frac{1}{2023})$

$=5(\frac{1}{2}-\frac{1}{2023})=\frac{10105}{4046}$

\(S=\dfrac{2^2}{1.2}+\dfrac{2^2}{2.3}+\dfrac{2^2}{3.4}+...+\dfrac{2^2}{2022.2023}\)

\(S=2^2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)\)

\(S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)

\(S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2023}\right)\)

\(S=2^2.\dfrac{2022}{2023}\)

\(S=\dfrac{2^2.2022}{2023}=\dfrac{8088}{2023}\)

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 

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

\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)=2023x\)

\(\Rightarrow2022x+\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}+\dfrac{1}{2022}-\dfrac{1}{2023}\right)=2023x\)\(\Rightarrow2022x-2023x=-\left(1-\dfrac{1}{2023}\right)\)

\(\Rightarrow-x=-\dfrac{2022}{2023}\Leftrightarrow x=\dfrac{2022}{2023}\)

(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