A=(1+2-3)+(-4+5+6-7)+(-8+9+10-11)+......(-2000+2001+2002-2003)

A=0+0....+0

A=0

Ta thấy 2-3-4=-5

            6-7-8=-9

           .............

           1998-1999-2000=-2001

=> 1+2-3-4+5+6-7-8+....-1999-2000+2001-2003=1-5+5-9+9-...-2001+2001+2002-2003

=> A= 1+2002-2003=0

Vậy A=0

\(=\left(1+2-3\right)+\left(-4+5+6-7\right)+...+\left(-2000+2001+2002-2003\right)\)

\(=0+0+0+...+0\)

\(=0\)

học tốt

S = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + 10 - ...... + 1998 - 1999 - 2000 + 2001 + 2002

S = 1 + (2 - 3 - 4 + 5 )+ (6 - 7 - 8 + 9) + (10 - ...... + (1998 - 1999 - 2000 + 2001) + 2002

S=1+0+0...+0+2002

S= 1+2002

S=2003

Lời giải:

$S=(1+2-3-4)+(5+6-7-8)+(9+10-11-12)+...+(1997+1998-1999-2000)+2001+2002$

$=\underbrace{(-4)+(-4)+....+(-4)}_{500}+2001+2002$

$=(-4).500+2001+2002=2003$

`S = 1+2-3-5+5+6-7-8+9+10-...+1998-1999-2000+2001+2002`

có :

`(2002 - 1) :1 +1 = 2002` ( số hạng)

`2002 : 4 = 500 (dư 2)`

`=(1+2-3-4)+(5+6-7-8)+(9+10-11-12)+...+(1997+1998-1999-2000)+2001+2002`

`=(-4)+(-4)+...+(-4) +2001 +2002`  có `500` só `-4`

`=500 .(-4) + 2001+ 2002`

`= (-2000)+2001+2002`

`=1+2002`

`=2003`

Ta có : 

\(A=\frac{3}{4.5}+\frac{3}{5.6}+\frac{3}{6.7}+...+\frac{3}{99.100}\)

\(A=3\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\right)\)

\(A=3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=3\left(\frac{1}{4}-\frac{1}{100}\right)\)

\(A=3.\frac{6}{25}\)

\(A=\frac{18}{25}\)

Vậy \(A=\frac{18}{25}\)

Chúc bạn học tốt ~ 

\(A=\frac{3}{4.5}+\frac{3}{5.6}+\frac{3}{6.7}+...+\frac{3}{99.100}\)

\(\Rightarrow A=3.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\right)\)

\(\Rightarrow A=3.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(\Rightarrow A=3.\left(\frac{1}{4}-\frac{1}{100}\right)=\frac{3.24}{100}\)

\(=\frac{3.4.6}{25.4}\)

\(\Rightarrow A=\frac{18}{25}\)

cám ơn 2 bạn nhiềuạ!

D = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... - 1999 - 2000 + 2001 + 2002 - 2003

D = ( 1 + 2 - 3 - 4 ) + ( 5 + 6 - 7 - 8 ) + ... + ( 1997 + 1998 - 1999 - 2000 ) + 2001 + 2002 - 2003

D = ( -4 ) + ( -4 ) + ... + ( -4 ) + ( 2001 + 2002 - 2003 )

D = ( -4 ) . 500 + 2000

D = -2000 + 2000

D = 0

D = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ............. - 1999 - 2000 + 2001 + 2002 - 2003

D = ( 1 + 2 - 3 - 4 ) + ( 5 + 6 - 7 - 8 ) + ............ + ( 1997 + 1998 - 1999 - 2000 ) + 2001 + 2002 - 2003

D = ( -4 ) + ( -4 ) + .............. + ( -4 ) + ( 2001 + 2002 - 2003 )

D = ( -4 ) . 500 + 2000 

D = -2000 + 2000

D = 0

h) -1 - 2 - 3 - 4 - 5 - ..... - 1999 - 2000 - 2001 - 2002
A= -(1 + 2 + 3+...+ 2002) có 2002 số hạng
A= -(2002+2001+2000+...+2+1)
2A=-(1+2002) .2002
A=-2003.1001 =-2005003

-1 - 2 - 3 - 4 - 5 - .... - 1999 - 2000 - 2001 - 2002

Ta có : -1 - 2 - 3 - 4 - 5 - .... - 1999 - 2000 - 2001 - 2002

          = - (1 + 2 + 3 + 4 + ... + 2000 + 2001 + 2002)  (có 2002 số )

         = - [(2002 + 1) . 2002 : 2] 

         = - 2005003