( 1999 x 1998 + 1998 + 1997 ) + ( 1 + 1/2 / 11/2 - 11/3 )

= [ 1998( 1999 + 1 ) + 1997 ] + ( 3/2 / 11/6 )

= ( 1998 x 2000 + 1997 ) + ( 3/2 . 6/11 ) 

= 3997997 + 9/11

= 3997997,818

6/2(3*2)=0,5=1/2

`D = 2000 - 1999 + 1998 -1997 + ... + 2 - 1`

`D = (2000-1999)+(1998-1997)+...+(2-1)`

`D = 1+1+...+1`

Ta có: `(2000-1) \div 1 + 1 = 1000`

`\rightarrow` Có `1000` hiệu tương tự

`\rightarrow` `D= 1+1+...+1 = 1000.`

Để ý: 2000 - 1999= 1

1998-1997=1

....

2-1=1

Có 1000 hiệu như vậy

D = 1000

     1/2000*1999 - 1/1999*1998 - 1/1998*1997 - ... - 1/2*1

  = 1/1999 - 1/2000 - (1/1998 - 1/1999) - (1/1997 - 1/1998) - ... - (1 - 1/2)

 = 1/1999 - 1/2000 - 1/1998 + 1/1999 - 1/1997 +1/1998 - .... - 1 + 1/2

 = 1/1999 + 1/1999 - 1/2000  - 1/1998 + 1/1998 - 1/1997 +1/1997 - .... - 1/2 +1/2 - 1

 = 1/1999 + 1/1999 - 1/2000 - 1 

 = 2/1999 - 1 - 1/2000 

= -1997/1999 - 1/2000

= -2000 - 1997/1997*2000

=-3997/3994000

= 

Tổng trên có: (2002 -1):1+1 =2002 (số hạng)

Ta có: 1+ (2 -3 -4 +5)+ (6 -7 -8 +9)+...+(1998 -1999 -2000 -2001)+ 2002   (có 500 nhóm, thừa 2 số)

       = 1 + 0 + 0 + ... + 0 + 2002

       = 2003

Chúc bạn học tốt.

cac ban tra loi giup mk voi 

\(D=\dfrac{1}{2000.1999}-\dfrac{1}{1999.1998}-\dfrac{1}{1998.1997}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(D=\dfrac{1}{1999.2000}-\left(\dfrac{1}{1998.1999}+\dfrac{1}{1997.1998}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)\)\(D=\dfrac{1}{1999.2000}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{1997.1998}+\dfrac{1}{1998.1999}+\dfrac{1}{1999.2000}\right)\)

\(D=\dfrac{1}{1999.2000}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{1997}-\dfrac{1}{1998}+\dfrac{1}{1998}-\dfrac{1}{1999}+\dfrac{1}{1999}-\dfrac{1}{2000}\right)\)\(D=\dfrac{1}{1999.2000}-\dfrac{1999}{2000}\)

\(1+2-3-4+5+6-7-8+...+1997+1998-1999+2000\)

\(=\left(1+2-3-4\right)+...+\left(1997+1998-1999-2000\right)\)

\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)

\(=\left(-4\right).500\)

\(=\left(-2000\right)\)

1+2-3-4+5+6-7-8+...+1997+1998-1999-2000

=(1+2-3-4)+...+(1997+1998-1999-2000)

=(-4)+(-4)+...+(-4)

=(-4)x500

=(-2000)

5¹⁹⁹⁸+5²⁰⁰⁰+5²⁰⁰²=5¹⁹⁹⁸x(1+2+4)=5¹⁹⁹⁸x7

5¹⁹⁹⁹+5¹⁹⁹⁷+5¹⁹⁹⁵=5¹⁹⁹⁵x(1+2+4)=5¹⁹⁹⁵x7

(5¹⁹⁹⁸+5²⁰⁰⁰+5²⁰⁰²):(5¹⁹⁹⁹+5¹⁹⁹⁷+5¹⁹⁹⁵)=(5¹⁹⁹⁸x7):(5¹⁹⁹⁵x7)=(5¹⁹⁹⁸:5¹⁹⁹⁵)x(7:7)=5³x1=125

Vậy (5¹⁹⁹⁸+5²⁰⁰⁰+5²⁰⁰²)+(5¹⁹⁹⁹+5¹⁹⁹⁷+5¹⁹⁹⁵)=15