A=1-3+5-7+....+2001-2003+2005

A=[(1-3)+(5-7)+.....+(2001-2003)]+2005

A=[(-2)+(-2)+....+(-2)]+2005

Vì từ 1 đến 2003 có: 1002 số hạng => có 501 cặp => có 501 số -2

A=(-2) x 501 +2005

A=-1002+2005

A=1003

A=1-3+5-7+...+2001-2003+2005

A=(1-3)+(5-7)+....+(2001-2003)+2005

A=(-2)+(-2)+...+(-2)+2005

A=(-2).501+2005

A=(-1002)+2005

A=1003

B=1-2-3+4+5-6-7+8+...+1993-1994

B=(1-2-3+4)+(5-6-7+8)+....+(1989-1990-1991+1992)+(1993-1994)

B=0+0+...+0+(-1)

B=(-1)

C=1+2-3-4+5+6-7-8+9+...+2002-2003-2004+2005+2006

C=(1+2-3-4)+(5+6-7-8)+....+(2001+2002-2003-2004)+(2005+2006)

C=(-4)+(-4)+....+(-4)+4011

C=(-4).501+4011

C=(-2004)+4011

C=2007

A=1-3+5-7+...+2001-2003+2005

A= (-2) + (-2) +....+(-2)  +2005

A= -2. 501 +2005

A= -1002 +2005

A= 1003

B=1-2-3+4+5-6-7+8+...+1993-1994

B= (1-2-3+4) + (5-6-7 +8) +.......+ (1989 - 1990 -1991 +1992)+1993-1994

B= 0 + 0+....+0+ 1993-1994

B= -1

C=1+2-3-4+5+6-7-8+9+...+2002-2003-2004+2005+2006

C= (1+2-3-4) + (5+6-7-8) +.....+(2001+2002 -2003 -2004) +2005+2006

C= -4. 501 + 2005 +2006

C= -2004+2005+2006

C= 2007

Lời giải:
a. 

$A=(1-3)+(5-7)+(9-11)+...+(2001-2003)+2005$

$=(-2)+(-2)+(-2)+...+(-2)+2005$

$=(-2).501+2005=-1002+2005=1003$

b.

$B=(1-2-3+4)+(5-6-7+8)+...+(1989-1990-1991+1992)+(1993-1994)$

$=0+0+....+0+(1993-1994)=0+(-1)=-1$

Mời bạn tham khảo các link sau: 

a),b),c):https://hoidap247.com/cau-hoi/214111

d):https://olm.vn/hoi-dap/detail/78449788871.html

giúp mik với

a) A=1-2-3+4+5-6-7+.....+1996+1997-1998-1999+2000

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

=0

b) B=1-3+5-7+....+2001-2003+2005

=(1-3)+(5-7)+...+(2001-2003)+2005

=-2.501+2005

=-1002+2005

=1003

c) C=1-2-3+4+5-6-7+8+.....+1993-1994-1995+1996+1997

=(1-2-3+4)+(5-6-7+8)+...+(1993-1994-1995+1996)+1997

=1997

d) D=1000+998+996+......+10-999-997-995-...-11

=(1000-999)+(998-997)+(996-995)+....+(12-11)+10

=1.495+10

=595

$A=1-$$3+5-7+...+2001-2003+2005$

$=\left(1-3\right)+\left(5-7\right)+...+\left(2001-2003\right)+2005$(Có 1002 cặp)

$=\left(-2\right).1002+2005$

$=-2004+2005$

$=1$