TÌm x biết
a) \(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+....+\dfrac{1}{x\left(x+1\right):2}=1\dfrac{2019}{2021}\)
Những câu hỏi liên quan
Tìm x biết : \(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+.....+\dfrac{2}{x\left(x+1\right)}=1\dfrac{2019}{2021}\)
\(1+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{2}{x\left(x+1\right)}=1\dfrac{2019}{2021}\)
\(\Leftrightarrow\dfrac{1}{\dfrac{1\cdot2}{2}}+\dfrac{1}{\dfrac{2\cdot3}{2}}+\dfrac{1}{\dfrac{3\cdot4}{2}}+...+\dfrac{1}{\dfrac{x\left(x+1\right)}{2}}=\dfrac{4040}{2021}\)
\(\Leftrightarrow\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{4040}{2021}\)
\(\Leftrightarrow2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{4040}{2021}\)
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2020}{2021}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{2020}{2021}\)
\(\Leftrightarrow\dfrac{x}{x+1}=\dfrac{2020}{2021}\)
\(\Leftrightarrow2021x=2020x+2020\Leftrightarrow x=2020\)
Vậy S = {2020}
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Cho mk hỏi :
tìm x,biết
a,\(-3\dfrac{1}{2}\)x -0,75-1,25x=\(\left(\dfrac{-1}{2}\right)^2:\dfrac{-3}{4}+\dfrac{1}{6}\)
b, \(\dfrac{-2}{3}-\left(\dfrac{x}{2}-75\%\right)=\left(\dfrac{3}{-4}-\dfrac{9}{8}\right)^2:\dfrac{-3}{32}-1\dfrac{1}{3}\)
a, \(\left(2x-1\right)\left(x+\dfrac{2}{3}\right)=0\)
b, \(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)
a)
`(2x-1)(x+2/3)=0`
\(< =>\left[{}\begin{matrix}2x-1=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
b)
\(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)
\(< =>\dfrac{x+4}{2019}+1+\dfrac{x+3}{2020}+1=\dfrac{x+2}{2021}+1+\dfrac{x+1}{2022}+1\)
\(< =>\dfrac{x+2023}{2019}+\dfrac{x+2023}{2020}=\dfrac{x+2023}{2021}+\dfrac{x+2023}{2022}\)
\(< =>\left(x+2023\right)\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\right)=0\)
\(< =>x+2023=0\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\ne0\right)\\ < =>x=-2023\)
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a) + Chia thành 2 trường hợp
- 2x - 1 = 0
2x = 0 + 1
2x = 1
x = 1 : 2
x = 0,5
- x + 2/3 = 0
x = 0 - 2/3
x = -2/3
vậy x = { 0,5 ; -2/3 }
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Xem thêm câu trả lời
8 tháng 3 2022 lúc 17:09
Tìm x:
\(\dfrac{1}{3}\) + \(\dfrac{1}{6}\)+ \(\dfrac{1}{10}\)+ ...+ \(\dfrac{1}{xx\left(x+1\right):2}\)= \(\dfrac{2017}{2019}\)
Mọ người giúp em với ạ! Em cảm ơn!
mn ghi giúp em chi tiết bài giải nx ạ!
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2) tìm x biết
a) \(\left(\dfrac{-3}{4}x+1\right):\dfrac{2}{3}=1\)
b) (x+3)=6
\(\left(-\dfrac{3}{4}x+1\right)\div\dfrac{2}{3}=1\)
\(-\dfrac{3}{4}x+1=1\times\dfrac{2}{3}\)
\(-\dfrac{3}{4}x+1=\dfrac{2}{3}\)
\(-\dfrac{3}{4}x=\dfrac{2}{3}-1\)
\(-\dfrac{3}{4}x=-\dfrac{1}{3}\)
\(x=-\dfrac{1}{3}\div\left(-\dfrac{3}{4}\right)\)
\(x=\dfrac{4}{9}\)
x+3=6
x=6-3
x=3
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4 tháng 5 2022 lúc 20:30
Câu 1: Thực hiện phép tính
a, \(40\dfrac{1}{4}:\dfrac{5}{7}-25\dfrac{1}{4}:\dfrac{5}{7}-\dfrac{1}{2021}\)
b, \(\left|\dfrac{-5}{9}\right|.\sqrt{81}-2021^0.\dfrac{16}{25}\)
Câu 2: Tìm x
\(3\left(x-\dfrac{1}{3}\right)-7\left(x+\dfrac{3}{7}\right)=-2x+\dfrac{1}{3}\)
1:
a: =7/5(40+1/4-25-1/4)-1/2021
=21-1/2021=42440/2021
b: =5/9*9-1*16/25=5-16/25=109/25
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Cho 3 số x, y, z TM: \(\left\{{}\begin{matrix}x+y+z=2017\\\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{2017}\end{matrix}\right.\)
Tính GTBT: \(P=\left(x^{2017}+y^{2017}\right)\left(y^{2019}+z^{2019}\right)\left(z^{2021}+x^{2021}\right)\)
11 tháng 5 2022 lúc 17:09
\(Tìm.x:\)
\(2x-\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right)...\left(1+\dfrac{1}{2019+2021}\right)=x+1\)
ét o ét .-.
quy luật lạ z:v lúc đầu nhân lúc sau cộng:v
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Tìm x:
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+......+\dfrac{2}{x\left(x+1\right)}=\dfrac{2017}{2019}\)
\(\dfrac{1}{3}+\dfrac{1}{6}+....+\dfrac{2}{x\left(x+1\right)}=\dfrac{2017}{2019}\\ \Rightarrow\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2017}{2019}\\ \Rightarrow2.\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2017}{2019}\\ \Rightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2017}{4038}\\ \Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2019}\\ \Rightarrow x=2018\)
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