a)0-1+2-3+4-5+6-7+......+2004-2005                                         

= ( 0 - 1 ) + ( 2 - 3 ) + ( 4 - 5 ) + ( 5 - 6 ) + ... + ( 2004 - 2005 )

= (-1) + (-1) + (-1) + (-1) + ... + (-1) ( có 1003 số -1 ( 2005 - 0 ) : 1 + 1 = 2006 số hạng=> có 1003 cặp=> có 1003 số -1 )

= (-1) . 1003

= -1003

b)1-3+5-7+9-11+......+2005-2007

     = ( 1-3 ) + ( 5-7 ) + ( 9 - 11 ) +......+ ( 2005-2007 )

     = (-2) +  (-2) +  (-2) +  (-2) + ...  + (-2)  ( có ( 2017 -1):2+1= 1009 số hạng, chỗ này lẻ mất rồi , ko chia ra cặp được nhé

đúng đó bạn ạ

Câu F có lộn đề ko bạn

a)    \(\frac{x+1}{4}-\frac{x+2}{5}+\frac{x+4}{7}-\frac{x+5}{8}+\frac{x+7}{10}-\frac{x+9}{12}=0\)

\(\Leftrightarrow\)\(\frac{x+1}{4}-1-\frac{x+2}{5}+1+\frac{x+4}{7}-1-\frac{x+5}{8}+1+\frac{x+7}{10}-1-\frac{x+9}{12}+1=0\)

\(\Leftrightarrow\)\(\frac{x-3}{4}-\frac{3-x}{5}+\frac{x-3}{7}-\frac{3-x}{8}+\frac{x+3}{10}-\frac{3-x}{12}=0\)

\(\Leftrightarrow\)\(\frac{x-3}{4}+\frac{x-3}{5}+\frac{x-3}{7}+\frac{x-3}{8}+\frac{x-3}{10}+\frac{x-3}{12}=0\)

\(\Leftrightarrow\)\(\left(x-3\right)\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}\right)=0\)

Vì   \(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}\ne0\)

\(\Rightarrow\)\(x-3=0\)

\(\Leftrightarrow\)\(x=3\)

Vậy...

b)   \(\frac{x}{2004}+\frac{x+1}{2005}+\frac{x+2}{2006}+\frac{x+3}{2007}=4\)

\(\Leftrightarrow\)\(\frac{x}{2004}-1+\frac{x+1}{2005}-1+\frac{x+2}{2006}-1+\frac{x+3}{2007}-1=0\)

\(\Leftrightarrow\)\(\frac{x-2004}{2004}+\frac{x-2004}{2005}+\frac{x-2004}{2006}+\frac{x-2004}{2007}=0\)

\(\Leftrightarrow\)\(\left(x-2004\right)\left(\frac{1}{2004}+\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}\right)=0\)

Vì   \(\frac{1}{2004}+\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}\ne0\)

\(\Rightarrow\)\(x-2004=0\)

\(\Leftrightarrow\)\(x=2004\)

Vậy...

A = 0-1 + 2-3 + 4-5 +...+ 2017-2018 

=> A = (-1) + (-1) + (-1)  +...+ (-1) (Có 1009 số hạng)

=> A = 1009.(-1)

=> A = -1009

B = 1-3+5-7+ 9-11+....+2005-2007 

=> B = (-2) + (-2) +(-2) +...+ (-2) (Có 502 số hạng)

=> B = 502.(-2)

=> B = -1004

C=1+2+3-4-5-6+7+8+9-10-11-12+.....+97+98+99-100-101-102

=> C = (1+2+3-4-5-6)+...+(97+98+99-100-101-102) (có 17 cặp số)

=> C = (-9) + (-9) +...+ (-9) (có 17 số hạng)

=> C = (-9).17

=> C = -153