(X+1)+(x+2)+(x+3)+.....+(x+2022)=2023
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I
5 tháng 1 2023
\(\dfrac{x+1}{2023}+\dfrac{x+2}{2022}=\dfrac{x+3}{2021}+\dfrac{x+4}{2020}\\ \Leftrightarrow\dfrac{x+1}{2023}+1+\dfrac{x+2}{2022}+1=\dfrac{x+3}{2021}+1+\dfrac{x+4}{2020}+1\\ \Leftrightarrow\dfrac{x+1+2023}{2023}+\dfrac{x+2+2022}{2022}-\dfrac{x+3+2021}{2021}-\dfrac{x+4+2020}{2020}=0\\ \Leftrightarrow\left(x+2024\right)\times\left(\dfrac{1}{2023}+\dfrac{1}{2022}-\dfrac{1}{2021}-\dfrac{1}{2020}\right)=0\\ \Rightarrow x+2024=0:\left(\dfrac{1}{2023}+\dfrac{1}{2022}-\dfrac{1}{2021}-\dfrac{1}{2020}\right)\\ \Rightarrow x+2024=0\\ \Rightarrow x=-2024\)
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
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{x.\left(x+1\right)}\)
= \(\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+...+\dfrac{x+1-x}{x.\left(x+1\right)}\)
= \(\dfrac{2}{1.2}-\dfrac{1}{1.2}+\dfrac{3}{2.3}-\dfrac{2}{2.3}+...+\dfrac{x+1}{x.\left(x+1\right)}-\dfrac{x}{x.\left(x+1\right)}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)
= \(1-\dfrac{1}{x+1}\) =\(\dfrac{2022}{2023}\)
= \(\dfrac{2023}{2023}-\dfrac{1}{x+1}=\dfrac{2022}{2023}\)
⇒ \(x+1=2023\)
\(x=2023-1=2022\)
AH
Akai Haruma
Giáo viên
30 tháng 3 2023
Bạn cần viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt hơn.
NN
3
17 tháng 9 2020
\(\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+3}{2017}=\frac{x-1}{2021}+\frac{x-2}{2022}+\frac{x-3}{2023}\)
\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)+\left(\frac{x+3}{2017}+1\right)=\left(\frac{x-1}{2021}+1\right)+\left(\frac{x-2}{2022}+1\right)+\left(\frac{x-3}{2023}+1\right)\)
\(\Leftrightarrow\left(\frac{x+1+2019}{2019}\right)+\left(\frac{x+2+2018}{2018}\right)+\left(\frac{x+3+2017}{2017}\right)=\left(\frac{x-1+2021}{2021}\right)+\left(\frac{x-2+2022}{2022}\right)+\left(\frac{x-3+2023}{2023}\right)\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}=\frac{x+2020}{2021}+\frac{x+2020}{2022}+\frac{x+2020}{2023}\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}-\frac{x+2020}{2021}-\frac{x+2020}{2022}-\frac{x+2020}{2023}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\right)=0\)
Vì \(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\ne0\)
=> x + 2020 = 0
=> x = -2020
B
17 tháng 9 2020
Bài làm :
Ta có :
\(\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+3}{2017}=\frac{x-1}{2021}+\frac{x-2}{2022}+\frac{x-3}{2023}\)
\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)+\left(\frac{x+3}{2017}+1\right)=\left(\frac{x-1}{2021}+1\right)+\left(\frac{x-2}{2022}+1\right)+\left(\frac{x-3}{2023}+1\right)\)
\(\Leftrightarrow\left(\frac{x+1+2019}{2019}\right)+\left(\frac{x+2+2018}{2018}\right)+\left(\frac{x+3+2017}{2017}\right)=\left(\frac{x-1+2021}{2021}\right)+\left(\frac{x-2+2022}{2022}\right)+\left(\frac{x-3+2023}{2023}\right)\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}=\frac{x+2020}{2021}+\frac{x+2020}{2022}+\frac{x+2020}{2023}\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}-\frac{x+2020}{2021}-\frac{x+2020}{2022}-\frac{x+2020}{2023}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\right)=0\)
\(\text{Vì : }\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\ne0\)
\(\Rightarrow x+2020=0\Leftrightarrow x=-2020\)
Vậy x=-2020
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
5 tháng 5 2023
\(\dfrac{2022\times2023-1}{2023\times2021+2022}\)
= \(\dfrac{\left(2021+1\right)\times2023-1}{2023\times2021+2022}\)
= \(\dfrac{2023\times2021+2023-1}{2023\times2021+2022}\)
= \(\dfrac{2023\times2021+2022}{2023\times2021+2022}\)
= 1
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
=>2022x+2022*2023/2=2023
=>2022x=-2043230
=>x=-1010,5