Question

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

Answer 1

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 

Answer 2

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 

Answer 3

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

Answer 4

= 2 x [1/6 + 1/12 + 1/20 +...+ 2/x[x+1] = 2003/2005

=1/2 x 3+ 1/3 x 4+ 1/4 x5 +...+ 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

vay la : 2004

Answer 5

nhân VT với 1/2 là được nha bn