Tìm số tự nhiên x, biết rằng:
1/3+1/6+1/10+...+2/x.(x+1)=2018/2019
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NL
2
XO
14 tháng 7 2021
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{2021}\)
<=> \(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)
<=> \(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)
<=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4042}\)
<=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{2042}\)
<=> \(\frac{1}{x+1}=\frac{1}{2021}\)
<=> x + 1 = 2021
<=> x = 2020
16 tháng 7 2021
Có phải là bình 6a3 học trường THCS Nguyễn Trãi đúng không
VQ
5
30 tháng 7 2020
Đề bạn thiếu 1 số \(x\) nữa đúng không?
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2021}\)
\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4042}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4042}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4042}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{2021}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2021}\)
\(\Rightarrow x+1=2021\)
\(\Rightarrow x=2020\)
Vậy \(x=2020\).
KN
30 tháng 7 2020
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2021}\)
\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4042}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4042}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4042}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2019}{4042}=\frac{1}{2021}\)
\(\Leftrightarrow x+1=2021\)
\(\Leftrightarrow x=2020\left(tm:x\in N\right)\)
5 tháng 10 2021
Bài 2:
Ta có: \(16x+40=10\cdot3^2+5\left(1+2+3\right)\)
\(\Leftrightarrow16x+40=90+30\)
\(\Leftrightarrow16x=80\)
hay x=5
NM
3
9 tháng 5 2017
Tìm số tự nhiên x biết
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\cdot\left(x+1\right):2}=\frac{2016}{2018}\)
7 tháng 10 2018
a)xét 2A =2+2^2+2^3+.....+2^2019
-A=1+2+2^2+...+2^2018
A=(2^2019)-1 <2^2019
b)theo câu a ta có A+1=2^2019-1+1=2^2019=2^(x+1)
2019=x+1 =>x=2018
ZV
3
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
12 tháng 5 2016
Đặt vế trái là A ta có:
\(\frac{A}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)
\(\frac{A}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\)
\(\frac{A}{2}=\frac{1}{2}-\frac{1}{x+1}\Rightarrow\frac{A}{2}=\frac{x+1-2}{2\left(x+1\right)}\Rightarrow A=\frac{x-1}{x+1}\)
\(\Rightarrow\frac{x-1}{x+1}=\frac{2007}{2009}\Leftrightarrow x=2003\)
29 tháng 11 2022
\(\frac{A}{2}=\frac{1}{2}-\frac{1}{x+1}\Rightarrow\frac{A}{2}=\frac{x+1-2}{2\left(x+1\right)}\Rightarrow...
QD
3
PQ
28 tháng 8 2015
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+..+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1999}{2001}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{1999}{2001}\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{2001}:2\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{2001}:2=\frac{1}{2001}\Rightarrow x+1=2001\Rightarrow x=2000\)
N
1
7 tháng 5 2015
ta có: 1/3 + 1/6 + ... + 2/x(x+1) = 2/2.3 + 2/3.4 +.......2/x(x+1) = 2(1/2.3 +1/3.4 +.....+1/x(x+1)) = 2.(1/2-1/3+1/3-1/4+....+1/x-1/(x+1))= 2.(1/2-1/(x+1)) = 1-2/(x+1)
giải 1-2/(x+1) = 2007/2009 ta được x=2008
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)
\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)
\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2018}{2019}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1009}{2019}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{4038}\)
\(\Rightarrow x+1=4038\)
\(\Rightarrow x=4037\)
Vậy \(x=4037\)
\(\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)
\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}+\frac{1}{x+1}\right)=\frac{2018}{2019}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1009}{2019}\)
\(\frac{1}{x+1}=\frac{1}{4038}\)
\(x=4037\)