Question

Tìm số tự nhiên x biết rằng:

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{2001}\)

Answer 1

Đặt: A= 1/3 +1/6+1/10+…+2/x(x+1)
A x 1/2 = 1/2.3 + 1/3.4 + 1/4.5 +…+1/x(x+1)
A x1/2 = 1/2-1/3+1/3-1/4+1/4-1/5+…..+1/x-1/(x+1)
A x 1/2 = 1/2 – 1/(x+1)
A = (1/2 -1/x+1) : 1/2
A = 1 – 2/(x+1)
Như vậy ta có: 1-2/(x+1) = 1999/2001
Hay: 2/(x+1) = 1-1999/2001
2/(x+1) = 2/2001
Vậy x = 2000

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Answer 2

Ta có: 

Answer 3

1/3+1/6+1/10+...+1/x(x+1)=1999/2001

1/2.[1/3+1/6+1/10+...+1/x(x+1)].2=1999/2001

[1/6+1/12+1/20+...+1/x(x+1)].2=1999/2001

[1/2.3+1/3.4+1/4.5+...+1/x.(x+1)].2=1999/2001

[1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1].2=1999/2001

(1/2-1/x+1).2=1999/2001

1/2-1/x+1=1999/2001:2=1999/2001.1/2=1999/4002

1/x+1=1/2-1999/4002

1/x+1=2001/4002-1999/4002==2/4002=1/2001

=>x+1=2001

=>x=2001-1=2000

Vậy x=2000

Answer 4

1/3 + 1/6 + 1/10 + ... + 1/x ( x +1 ) = 1999/2001

=> 2/6 + 2/12 + 2/20 + .. + 2/ x ( x+1 )= 1999/2001

=> 2 ( 1/6 + 1/12 + ..+ 1/ x ( x +1 ) ) = 1999/2001

=> 2 ( 1/2 x 3 + 1/ 3 x 4 + ... + 1/x ( x + 1 ) ) = 1999/2001

=> 2 ( 1/2 - 1/3 + 1/3 - 1/4 + .. + 1/x - 1/ ( x+1 ) ) = 1999/2001

=> 2 ( 1/2 - 1/ ( x+ 1 ) ) = 1999/2001 => 1/2 - 1/ ( x + 1 ) = 1999/2001 : 2

=>  1/ x + 1 = 1/2 - 1999/2001 : 2 = 1/2001 => x + 1 = 2001 => x = 2000

Answer 5

Đặt: A= 1/3 +1/6+1/10+…+2/x(x+1)

A x 1/2 = 1/2.3 + 1/3.4 + 1/4.5 +…+1/x(x+1)

A x1/2 = 1/2-1/3+1/3-1/4+1/4-1/5+…..+1/x-1/(x+1)

x 1/2 = 1/2 – 1/(x+1) 

Answer 6

x=2000 ai k mk mk kb

Answer 7

x=2000 ak

Answer 8

1/3+1/6+1/10+...+1/x(x+1)=1999/2001

1/2.[1/3+1/6+1/10+...+1/x(x+1)].2=1999/2001

[1/6+1/12+1/20+...+1/x(x+1)].2=1999/2001

[1/2.3+1/3.4+1/4.5+...+1/x.(x+1)].2=1999/2001

[1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1].2=1999/2001

(1/2-1/x+1).2=1999/2001

1/2-1/x+1=1999/2001:2=1999/2001.1/2=1999/4002

1/x+1=1/2-1999/4002

1/x+1=2001/4002-1999/4002==2/4002=1/2001

=>x+1=2001

=>x=2001-1=2000

Vậy x=2000