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

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

\(\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{4008}{2005}\)

\(2.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{x\left(x+1\right)}\right)=\frac{4008}{2005}\)

\(2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{4008}{2005}\)

\(=>2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{4008}{2005}\)

\(2.\left(1-\frac{1}{x+1}\right)=\frac{4008}{2005}\)

=> \(1-\frac{1}{x+1}=\frac{4008}{2005}:2=\frac{2004}{2005}\)

\(\frac{1}{x+1}=1-\frac{2004}{2005}=\frac{1}{2005}\)

=>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

Ta có : 

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

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

\(\Leftrightarrow\)\(1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow\)\(1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)

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

\(\Leftrightarrow\)\(1+1-\frac{2}{x+1}=\frac{2003}{2005}\)

\(\Leftrightarrow\)\(\frac{2}{x+1}=2-\frac{2003}{2005}\)

\(\Leftrightarrow\)\(\frac{2}{x+1}=\frac{2007}{2005}\)

\(\Leftrightarrow\)\(x+1=2:\frac{2007}{2005}\)

\(\Leftrightarrow\)\(x+1=\frac{4010}{2007}\)

\(\Leftrightarrow\)\(x=\frac{4010}{2007}-1\)

\(\Leftrightarrow\)\(x=\frac{2003}{2007}\)

Vậy \(x=\frac{2003}{2007}\)

Chúc bạn học tốt ~ 

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

    1/3+1/6+...+1/x(x+1)=2003/2005

   1/2(1/3+1/6+..+1/x(x+1)=2003/4010

 1/6+1/12+...+1/x(x+1)=2003/4010

  1/2*3+1/3*4+...+1/x(x+1)=2003/4010

   1/2-1/3+1/3-1/4+...+1/x-1/x+1=2003/4010

 1/2-1/x+1=2003/4010

         1/x+1=1/2005

           x+1=2005

          x=2004

lộn

giải hông ra

%$&%$^%&*^%$%#$$%^%&^$%^#$^#^%&%^$%@#&^^%^&%%#%$#$#%$E%^$%^#$#@%@^#$%^E%^E$#&^^#$@#%^%^*&%#&(*($@$#(*^&^&*%$%#$#*%^^%$%^#^%#^*^>>>>>>>>>.............................................................................................. . mình không biết

\(\dfrac{2003}{20025}\) hay sao ý

GIẢI CHI TIẾT GIÙM MÌNH !!!!!