Tìm số tự nhiên x, biết:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2017}{2019}\)
Tìm x biết
a) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
b) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~
\(b)\) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(1-\frac{2}{x+1}=\frac{2017}{2019}\)
\(\Leftrightarrow\)\(\frac{2}{x+1}=1-\frac{2017}{2019}\)
\(\Leftrightarrow\)\(\frac{2}{x+1}=\frac{2}{2019}\)
\(\Leftrightarrow\)\(x+1=2019\)
\(\Leftrightarrow\)\(x=2019-1\)
\(\Leftrightarrow\)\(x=2018\)
Vậy \(x=2018\)
Chúc bạn học tốt ~
Tìm số tự nhiên x , biết rằng :
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)
1/3 + 1/6 + 1/10 + .......+2/x(x + 1) = 2015/2017
=> 2/2.3 +2/3.4 + 2/4.5 +........+ 2/x(x+1) =2015/2017
=> 2. [1/2.3 + 1/3.4 +1/4.5+....+1/x(x+1) ] = 2015/2017
=> 2. [ 1/2+ (-1/3 + 1/3) + (-1 /4 +1/4)+ -1/5 +.......+ 1/x + -1/x+1]
=> (1/2 + -1/x+1) .2 =2015/2017
=> 1/2 + -1/x+1 = 2015/2017 :2 = 2015/2017 . 1/2 =2015/4034.
=> -1/x+1 = 2015/4034 -1/2 = 2015/4034 -2017/4034 = -1/2017
=> -1/x+1 = -1/2017
=>x+1=2017
=> x= 2016
Tìm x biết
a) \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
b) \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2017}{2019}\)
\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(\Rightarrow x=305\)
a) \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(\Leftrightarrow x=308-3\)
\(\Leftrightarrow x=305\)
Vậy \(x=305\)
Tìm x biết :
a) \(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)(1)
Gọi B = \(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}\)
\(\Rightarrow3A=3\left(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}\right)\)
\(\Rightarrow3A=\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{x\left(x+3\right)}\)
\(\Rightarrow3A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\)
\(\Rightarrow3A=\frac{1}{5}-\frac{1}{x+3}\)
\(\Rightarrow A=\left(\frac{1}{5}-\frac{1}{x+3}\right):3\)(2)
Thay (2) vào (1) , ta có : \(\left(\frac{1}{5}-\frac{1}{x+3}\right):3=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(\Rightarrow x=305\)
Vậy x=305
Tương tự như câu b) nha
Chúc bạn học tốt!! nhớ mik nha
tìm số tự nhiên x
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.............+\frac{2.x}{x.\left(x+1\right)}=\frac{2005}{2017}\)
mình cần gấp
Tìm số tự nhiên x biết:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{2}{x.\left(x+1\right)}=\frac{1999}{2001}\)
\(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{2}{x.\left(x+1\right)}\)
Đặt A=1/3+1/6+1/10+...+2/x*(x+1)
1/2A=1/3*2+1/6*2+1/10*2+...+2/2*x*(x+1)
1/2A=1/6+1/12+1/20+...+1/x*(x+1)
1/2A=1/2*3+1/3*4+1/4*5+...+1/x*(x+1)
1/2A=1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/(x+1)
1/2A=1/2-1/x+1
A=(1/2-1/x+1):1/2
A=1-2/x+1
Ta có A=1999/2001
Hay 1-2/x+1=1999/2001
2/x+1=1-1999/2001
2/x+1=2/2001
=>x+1=2001
=>x=2000
Cho A = 1/3+1/6+1/10+...+2/x(x+1)
1/2A= 1/3.2+1/6.2+1/10.2+...+2/x(x+1)2
1/2A= 1/6+1/12+1/20+...+1/x(x+1)
1/2A= 1/2.3+1/3.4+1/4.5+...+1/x(x+1)
1/2A= 1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1
1/2A= 1/2-1/x+1
A = (1/2-1/x+1)/1/2
A = 1-2/x+1
Mà A=1999/2001
=> 1-2/x+1= 1999/2001
2/x+1= 1-1999/2001
2/x+1= 2/2001
=>x+1=2001
=>x = 2000
Đặt N=1/10+1/15+1/21+...+2/x*(x+1)
1/2N=1/20+1/30+1/42+...+1/x*(x+1)
1/2N=1/4*5+1/5*6+1/6*7+...+1/x*(x+1)
1/2N=1/4-1/5+1/5-1/6+1/6-1/7+...+1/x-1/x+1
1/2N=1/4-1/x+1
N=(1/4-1/x+1):1/2
N=1/2-2/x+1
Thiếu đề
Tìm số tự nhiên x biết \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2003}{2005}\)
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2003}{2005}\)
\(\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{4008}{2005}\)
\(2.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{x\left(x+1\right)}\right)=\frac{4008}{2005}\)
\(2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{4008}{2005}\)
\(=>2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{4008}{2005}\)
\(2.\left(1-\frac{1}{x+1}\right)=\frac{4008}{2005}\)
=> \(1-\frac{1}{x+1}=\frac{4008}{2005}:2=\frac{2004}{2005}\)
\(\frac{1}{x+1}=1-\frac{2004}{2005}=\frac{1}{2005}\)
=>x+1=2005
=>x=2004
1/3 + 1/6 + 1/10 +...+ 2/x(x+1) = 2014/2015
Đ/A là 2004
chúc đồng chí Chế Minh Hải học tốt
Tìm số tự nhiên x biết rằng : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)
\(\Rightarrow2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{2009}\div2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2007}{4018}\)
\(\Rightarrow\frac{1}{x+1}=\frac{2}{4018}=\frac{1}{2009}\)
\(\Rightarrow x+1=2009\)
\(\Rightarrow x=2008\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
=>\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2007}{4018}\)(nhân cả hai vế với \(\frac{1}{2}\))
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)= \(\frac{2007}{4018}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2007}{4018}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)
\(\frac{1}{x+1}\)=\(\frac{1}{2}-\frac{2007}{4018}\)
\(\frac{1}{x+1}=\frac{1}{2009}\)
x+1=2009
x=2009-1=2008
Vậy x bằng 2008
tìm số tự nhiên x biết rằng
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{2}{x\left(x+1\right)}=\frac{2000}{2002}\)
Ta có: 1/3+1/6+1/10+...+2/x*(x+1)
=2/6+2/12+2/20+...+2/x*(x+1)
=2/2*3+2/3*4+2/4*5+...+2/x*(x+1)
=2*(1/2*3+1/3*4+1/4*5+...+1/x*(x+1))
=2*(1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1)
=2*(1/2-1/x+1)=2000/2002
=>1/2-1/x+1=2000/2002:2
=>1/2-1/x+1=500/1001
=>1/x+1=1/2-500/1001
=>1/x+1=1/2002
=>x+1=2002
=>x=2002-1
=>x=2001 thuộc N
Vậy x=2001
*Mình ko biết ấn dấu phân số với dấu nhân ở đâu, bạn thông cảm nhé!
cuc si lau la ong be lac
co nhieu cau tuong tu ban tham tu khao nhe
Tìm số tự nhiên x biết: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\)