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

a, \(\frac{4}{7}x\frac{5}{7}+\frac{3}{7}x\frac{5}{6}\)

= ( \(\frac{4}{7}+\frac{3}{7}\)) x \(\frac{5}{7}x\frac{5}{6}\)

= 1 x \(\frac{5}{7}x\frac{5}{6}\)

= \(\frac{5}{7}x\frac{5}{6}\)

= \(\frac{25}{42}\)

Tương tự mấy câu sau cũng làm như thế này

a) Theo thứ tự từ bé đến lớn là: 1/5 ; 3/5  ; 4/5 ; 9/7

b) Theo thứ tự từ bé đến lớn là: 1/2005 ; 7/2005 ; 8/2005 ; 2006/2001

c) Theo thứ tự từ bé đến lớn là: 3/7 ; 29/48 ;294/343 ; 5/4

\(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+....+\frac{1}{2003\cdot2005}\)

\(2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{2003\cdot2005}\)

\(2A=\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+....+\frac{2005-2003}{2003\cdot2005}\)

\(2A=\left(\frac{3}{1\cdot3}-\frac{1}{1\cdot3}\right)+\left(\frac{5}{3\cdot5}-\frac{3}{3\cdot5}\right)+\left(\frac{7}{5\cdot7}-\frac{5}{5\cdot7}\right)+....+\left(\frac{2005}{2003\cdot2005}-\frac{2003}{2003\cdot005}\right)\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2003}-\frac{1}{2005}\)

\(2A=1-\frac{1}{2005}\)

\(2A=\frac{2004}{2005}\)

\(A=\frac{2004}{2005}\div2\)

\(A=\frac{2004}{4010}\)

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\)

( GẠCH BỎ CÁC PHÂN SỐ GIỐNG NHAU )

\(=1-\frac{1}{2005}\)

\(=\frac{2004}{2005}\)

CHÚC BN HỌC TỐT!!!

Mấy bạn làm sai rồi nhé =)))  Mình từng làm bài này rồi nên biết 

Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\)

\(\Leftrightarrow A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\right)\)

\(\Leftrightarrow A=\frac{1}{2}\left(1-\frac{1}{2005}\right)\)

\(\Leftrightarrow A=\frac{1}{2}.\frac{2004}{2005}=\frac{1002}{2005}\)

Đặt 

\(S=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2003+2005}\)

\(2S=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2003+2005}\)

\(2S=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2003}-\frac{1}{2005}\)

\(2S=\frac{1}{1}-\frac{1}{2005}=\frac{2004}{2005}\)

\(S=\frac{2004}{2005}.\frac{1}{2}=\frac{1002}{2005}\)

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

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

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

= ( -2 ) . 501 + 2005

= ( -1002 ) + 2005

= 1003 

~ Hok tốt ~

1 - 3 + 5 - 7 + ... + 2005 - 2007 + 2009

= ( 1 - 3 ) + ( 5 - 7 ) + ... + ( 2005 - 2007 ) + 2009

= ( -2 ) + ( -2 ) + ... + ( -2 ) + 2009

= ( -2 ) . 502 + 2009

= ( -1004 ) + 2009

= 1005

~ Hok tốt ~

giúp em nốt câu này

1-2+3-4+5-6+...+99-100+101

`a)17/7-3/8+4/7-3/5`

`=(17/7+4/7)-(3/8+3/5)`

`=21/7-(15/40+24/40)=3-39/40=120/40-39/40=81/40`

`b)15/11xx2001/2005-3/11xx2001/2005-1/11xx2001/2005`

`=2001/2005xx(15/11-3/11-1/11)`

`=2001/2005xx11/11`

`=2001/2005`