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

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!

uk mình cảm ơn bạn rất nhiều 

cuc si lau la ong be lac

co nhieu cau tuong tu ban tham tu  khao nhe

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

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}=\frac{2000}{2002}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{x\left(x+1\right)}=\frac{2000}{2002}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}\right)=\frac{2000}{2002}\)

\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2000}{2002}:2=\frac{1000}{2002}\)

=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{1000}{2002}=\frac{1}{2002}\)

=> x + 1 = 2002 

=> x = 2002 - 1 

=> x = 2001

Bạn Hồ Thu Giang làm rất tốt!

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2000}{2002}\)

\(\Rightarrow\) \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2000}{2002}\)

\(\Rightarrow\) \(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2000}{2002}\)

\(\Rightarrow\) \(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)

\(\Rightarrow\) \(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)

\(\Rightarrow\) \(\frac{1}{2}-\frac{1}{x+1}=\frac{500}{1001}\)

\(\Rightarrow\) \(\frac{1}{x+1}=\frac{1}{2002}\)

\(\Rightarrow\) \(x+1=2002\) \(\Rightarrow\) \(x=2001\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2000}{2002}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}\)=\(\frac{2000}{2002}\)

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)

\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{505}{1011}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1010}{1011}\)

=>1/x+1=-1009/2022

=>x+1=-2022/1009

hay x=-3031/1009

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