2)

đặt a= 1+2-3-4+5+6-........+2002-2003-2004+2005+2006

Biểu thức a có (2006-1)/1+1=2006(số hạng)

Nhóm 4 số hạng vào một nhóm ta có 2006 / 4= 501 dư 2 số hạng để ra một số đầu và một số cuối

a= 1+(2-3-4+5)+(6-7-8+9)-.........+(2002-2003-2004+2005) + 2006

a=1+0+0+......+0+2006

a=1+2006

a=2007 

vậy a = 2007

1/3 + 1/6 + 1/10 + ... + 2/x(x+1) = 2003/2005

2 × ( 1/6 + 1/12 + 1/20 + ... + 1/x(x+1) = 2003/2005

 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/x(x+1) = 2003/2005 : 2

1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1 = 2003/2005 × 1/2

1/2 - 1/x+1 = 2003/4010

1/x+1 = 1/2 - 2003/4010

1/x+1 = 2005/4010 - 2003/4010

1/x+1 = 1/2005

=> x+1 = 2005

=> x = 2004

Vậy x = 2004

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4010}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4010}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2005}\)

\(\Leftrightarrow x+1=2005\)

\(\Leftrightarrow x=2004\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4010}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4010}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2005}\)

\(\Leftrightarrow x=2004\)

1/3 + 1/6 + 1/10 + ... + 2/x(x+1) = 2003/2005

2 × ( 1/6 + 1/12 + 1/20 + ... + 1/x(x+1) = 2003/2005

 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/x(x+1) = 2003/2005 : 2

1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1 = 2003/2005 × 1/2

1/2 - 1/x+1 = 2003/4010

1/x+1 = 1/2 - 2003/4010

1/x+1 = 2005/4010 - 2003/4010

1/x+1 = 1/2005

=> x+1 = 2005

=> x = 2004

Vậy x = 2004

 ai tích mk tích lại cho 

1/3 + 1/6 + 1/10 + ... + 2/x(x+1) = 2003/2005

2 × ( 1/6 + 1/12 + 1/20 + ... + 1/x(x+1) = 2003/2005

 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/x(x+1) = 2003/2005 : 2

1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1 = 2003/2005 × 1/2

1/2 - 1/x+1 = 2003/4010

1/x+1 = 1/2 - 2003/4010

1/x+1 = 2005/4010 - 2003/4010

1/x+1 = 1/2005

=> x+1 = 2005

=> x = 2004

Vậy x = 2004

 ai tích mk tích lại cho 

1/3 + 1/6 + 1/10 + ... + 2/x(x+1) = 2003/2005

2 x (1/6 + 1/12 + 1/20 + ... + 1/x(x+1) = 2003/2005

 1/2x3 + 1/3x4 + 1/4x5 + ... + 1/x(x+1) = 2003/2005 : 2

1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1 = 2003/2005 x 1/2

1/2 - 1/x+1 = 2003/4010

1/x+1 = 1/2 - 2003/4010

1/x+1 = 2005/4010 - 2003/4010

1/x+1 = 1/2005

=> x+1 = 2005

=> x=2004

Vậy x = 2004

Gọi dãy trên là A

A= 2/6+2/12+2/20+...+2/x(x+1)  (mình không chép lại đề bài nhé)

  = 2(1/6+1/12+1/20+...+1/x(x+1)

  = 2(1/2.3+1/3.4+1/4.5+...+1/x(x+1)

  = 2(1/2-1/x+1)

  2(1/2-1/x+1)=2003/2005

     1/2-1/x+1 =2003/2005:2 ( tự làm tiếp nhé)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{x\left(x+1\right):2}=\frac{2001}{2003}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{2003}:2=\frac{2001}{4006}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}=\frac{1}{2003}\)

=> x+1 = 2003

=> x = 2003 - 1

=> x = 2002

thank you very much

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)

\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4010}\)

\(\Leftrightarrow\frac{x+1-2}{2\left(x+1\right)}=\frac{2003}{4010}\)

\(\Leftrightarrow2003.2\left(x+1\right)=4010\left(x-1\right)\)

\(\Leftrightarrow4006x+4006=4010x-4010\)

\(\Leftrightarrow-4x=-8016\)

\(\Leftrightarrow x=2004\)

Vậy x = 2004

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2003}{2005}\)

\(\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}\right).\frac{1}{2}=\frac{2003}{2005}.\frac{1}{2}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{2}{x.\left(x+1\right).2}=\frac{2003}{4020}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2003}{4020}\)

\(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{\left(x+1\right)-x}{x.\left(x+1\right)}=\frac{2003}{4020}\)

\(\frac{3}{2.3}-\frac{2}{2.3}+\frac{4}{3.4}-\frac{3}{3.4}+...+\frac{x+1}{\left(x+1\right).x}-\frac{x}{\left(x+1\right).x}=\frac{2003}{4020}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{\left(x+1\right)}=\frac{2003}{4020}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4020}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4020}=\frac{7}{4020}\)

\(\frac{7}{\left(x+1\right).7}=\frac{7}{4020}\)

\(\left(x+1\right).7=4020\)

\(\Rightarrow x=....\)