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15 tháng 12 2022
=>2022x+2022*2023/2=2023
=>2022x=-2043230
=>x=-1010,5
TT
0
DT
2
S
23 tháng 9 2023
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2022}{2023}\)
\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2022}{2023}\)
\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{2022}{2023}\)
\(\Rightarrow\dfrac{1}{x+1}=1-\dfrac{2022}{2023}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2023}\)
\(\Rightarrow x+1=2023\)
\(\Rightarrow x=2022\)
Vậy x = 2022
#kễnh
23 tháng 9 2023
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{x.\left(x+1\right)}\)
= \(\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+...+\dfrac{x+1-x}{x.\left(x+1\right)}\)
= \(\dfrac{2}{1.2}-\dfrac{1}{1.2}+\dfrac{3}{2.3}-\dfrac{2}{2.3}+...+\dfrac{x+1}{x.\left(x+1\right)}-\dfrac{x}{x.\left(x+1\right)}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)
= \(1-\dfrac{1}{x+1}\) =\(\dfrac{2022}{2023}\)
= \(\dfrac{2023}{2023}-\dfrac{1}{x+1}=\dfrac{2022}{2023}\)
⇒ \(x+1=2023\)
\(x=2023-1=2022\)
HK
16 tháng 4 2023
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = (\(\dfrac{2021}{2}+1\))+(\(\dfrac{2020}{3}+1\))+....+(\(\dfrac{1}{2022}+1\))
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = \(\dfrac{2023}{2}\)+\(\dfrac{2023}{3}\)+....+ \(\dfrac{2023}{2022}\)
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = 2023.( \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\))
vậy x= 2023
PN
1
NA
9 tháng 5 2022
`2x-15=-25`
`2x=-10`
`x=-5`
___________
`3/5<x/10<4/5`
`3/5=(3xx10)/(5xx10)=30/50`
`x/10=(5x)/(10xx5)=(5x)/50`
`4/5=(4xx10)/(5xx10)=40/50`
`=>30/50<(5x)/50<40/50`
`=>30<5x<40`
`=>x=7`
DB
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
6 tháng 1
\(3B=1.3^2+2.3^3+3.3^4+...+2022.3^{2023}+2023.3^{2024}\)
\(2B=3B-B=-3-3^2-3^3-...-3^{2023}+2023.3^{2024}\)
\(2B=2023.3^{2024}-\left(3+3^2+3^3+...+3^{2023}\right)\)
Đặt
\(C=3+3^2+3^3+...+3^{2023}\)
\(3C=3^2+3^3+3^4+...+3^{2024}\)
\(2C=3C-C=3^{2024}-3\Rightarrow C=\dfrac{3^{2024}-3}{2}\)
\(\Rightarrow2B=2023.3^{2024}-\dfrac{3^{2024}-3}{2}=\)
\(=\dfrac{2.2023.3^{2024}-3^{2024}+3}{2}=\dfrac{4045.3^{2024}+3}{2}\)
\(\Rightarrow B=\dfrac{4045.3^{2024}+3}{4}\)