tìm số tự nhiên x biết 1/3+1/6+1/10+...+2/x(x+1)=2000/2002
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B
1
17 tháng 5 2016
Ta có: \(A=\frac{1}{3}+\frac{1}{6}+......+\frac{2}{x.\left(x+1\right)}=\frac{2000}{2002}\)
\(A=\frac{1}{6}+\frac{1}{12}+......+\frac{1}{x.\left(x+1\right)}=\frac{2000}{2002}.\frac{1}{2}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x.\left(x+1\right)}=\frac{2000}{4004}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2000}{4004}\)
\(A=\frac{1}{2}-\frac{1}{x+1}=\frac{2000}{4004}\)
\(A=\frac{1}{x+1}=\frac{1}{2}-\frac{2000}{4004}\)
\(A=\frac{1}{x+1}=\frac{1}{2002}\)
\(x+1=2002\)
nên \(x=2002-1=2001\)
Vậy x = 2001
8 tháng 4 2018
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!
B
1
18 tháng 5 2016
Đặt A=1/3+1/6+1/10+...+1/(x(x+1))
=> A=2/6+2/12+2/20+...+2/2(x(x+1))
=> 1/2A=1/(2.3)+1/(3.4)+...+1/2(x(x+1))
=> 1/2A=1/2-1/3+1/3-1/4+...+1/x-1/(x+1)
=> 1/2A=1/2-1/(x+1). Vì A=2000/2002=1000/1001=> 1/2A=500/1001
=> 1/2-1/(x+1)=500/1001
=> 1/(x+1)=1001/2002-1000/2002
=> 1/(x+1)=1/2002
=> x+1=2002
=> x=2001
TM
2
HT
25 tháng 7 2015
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}=\frac{2000}{2002}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{x\left(x+1\right)}=\frac{2000}{2002}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}\right)=\frac{2000}{2002}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2000}{2002}:2=\frac{1000}{2002}\)
=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{1000}{2002}=\frac{1}{2002}\)
=> x + 1 = 2002
=> x = 2002 - 1
=> x = 2001
PT
Phạm Trần Hoàng Anh
CTVHS
7 tháng 7 2020
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2000}{2002}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}\)=\(\frac{2000}{2002}\)
2.(\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\))=\(\frac{2000}{2002}\)
2.\(\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2000}{2002}\)
2.(\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)) = \(\frac{2000}{2002}\)
2.\(\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1000}{2002}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{1000}{2002}\)
\(\frac{1}{x+1}=\frac{1}{2002}\)
2002.1 = (x+1).1
2002 = x+1
x=2001 (T/M)
MP
7 tháng 7 2020
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2000}{2002}\)
\(\Rightarrow\) \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2000}{2002}\)
\(\Rightarrow\) \(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2000}{2002}\)
\(\Rightarrow\) \(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)
\(\Rightarrow\) \(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)
\(\Rightarrow\) \(\frac{1}{2}-\frac{1}{x+1}=\frac{500}{1001}\)
\(\Rightarrow\) \(\frac{1}{x+1}=\frac{1}{2002}\)
\(\Rightarrow\) \(x+1=2002\) \(\Rightarrow\) \(x=2001\)
MN
1
11 tháng 2 2022
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{505}{1011}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1010}{1011}\)
=>1/x+1=-1009/2022
=>x+1=-2022/1009
hay x=-3031/1009
TQ
1
7 tháng 4 2023
Ta có:
1/3 + 1/6 + 1/10 + ... + 1/x(x+1):2 = 2001/2003
=> 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 2001/2003
=> 2 [1/6 + 1/12 + 1/20 + ... + 1/x(x+1)] = 2001/2003
=> 2 [1/2x3 + 1/3x4 + 1/4x5 + ... + 1/x+(x+1)] = 2001/2003
=> 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1= 2001/2003 : 2
=> 1/2 - 1/x+1 = 2001/4006
=> 1/x+1 = 1/2 - 2001/4006 = 1/2003
=> x+1 = 2003 = 2002 + 1
=>x = 2002
NM
0