Phải cho đề bài chứ. 

(X -10/1994 -1) + (X-8/1996 - 1) + (X-6/1998 - 1)+ (X-4/2000 - 1) + (X-2/2002 - 1) = (X-2002/2 - 1) + (X-2000/4 - 1) + (X-1998/6 - 1) + (X-1996/8 - 1) + (X-1994/10 - 1) 

=> x-2004/1994 + x-2004/1996 + x-2004/1998 + x-2004/2000 + x-2004/2002 = x-2004/2 + x-2004/4 + x-2004/6 + x-2004/8 + x-2004/1994 

=> x-2004/1994 + x-2004/1996 + x-2004/1998 + x-2004/2000 + x-2004/2002 - x-2004/2 -  x-2004/4 - x-2004/6 - x-2004/8 -  x-2004/1994 = 0 

=>  (x - 2004)(1/994 + 1/1996 + 1/1998 + 1/2000 + 1/2002 + 1/2 +  1/4 + 1/6 + 1/8) = 0 

Mà  (1/994 + 1/1996 + 1/1998 + 1/2000 + 1/2002 + 1/2 +  1/4 + 1/6 + 1/8) \(\ne\)0 

=> x - 2004 = 0 

=> x = 2004 

Vậy x = 2004 

Sửa (x  - 2004)  (1/1994 + 1/1996 + 1/1998/+ 1/2000 + 1/2002 + 1/2 + 1/4 +  1/6+ 1/8 + 1/10) 

 = (x - 2004) (1/1994 + 1/1996 + 1/1998/+ 1/2000 + 1/2002 - 1/2 -1/4 -  1/6 -  1/8 - 1/10) 

 :)))

\(\left(\frac{x-10}{1994}-1\right)\)+\(\left(\frac{x-8}{1996}-1\right)\)+\(\left(\frac{x-6}{1998}-1\right)\)+\(\left(\frac{x-4}{2000}-1\right)\)+\(\left(\frac{x-2}{2002}-1\right)\)=\(\left(\frac{x-2002}{2}-1\right)\)+\(\left(\frac{x-2000}{4}-1\right)\)+\(\left(\frac{x-1998}{6}-1\right)\)+\(\left(\frac{x-1996}{8}-1\right)\)+\(\left(\frac{x-1994}{10}-1\right)\)

suy ra \(\frac{x-2004}{1994}\)+\(\frac{x-2004}{1996}\)+\(\frac{x-2004}{1998}\)+\(\frac{x-2004}{2000}\)+\(\frac{x-2004}{2002}\)=\(\frac{x-2004}{2}\)+\(\frac{x-2004}{4}\)+\(\frac{x-2004}{6}\)+\(\frac{x-2004}{8}\)+\(\frac{x-2004}{10}\)

suy ra  \(\frac{x-2004}{1994}\)+\(\frac{x-2004}{1996}\)+\(\frac{x-2004}{1998}\)+\(\frac{x-2004}{2000}\)+\(\frac{x-2004}{2002}\)- \(\frac{x-2004}{2}\)- \(\frac{x-2004}{4}\)- \(\frac{x-2004}{6}\)- \(\frac{x-2004}{8}\)- \(\frac{x-2004}{10}\)=0

suy ra (x-2004) . ( \(\frac{1}{1994}\)+\(\frac{1}{1996}\)+\(\frac{1}{1998}\)+\(\frac{1}{2000}\)+\(\frac{1}{2002}\)-\(\frac{1}{2}\)-\(\frac{1}{4}\)-\(\frac{1}{6}\)- \(\frac{1}{8}\)- \(\frac{1}{10}\))=0

Vì  \(\frac{1}{1994}\)+\(\frac{1}{1996}\)+\(\frac{1}{1998}\)+\(\frac{1}{2000}\)+\(\frac{1}{2002}\)-\(\frac{1}{2}\)-\(\frac{1}{4}\)-\(\frac{1}{6}\)- \(\frac{1}{8}\)- \(\frac{1}{10}\) khác 0

nên x-2004=0 suy ra x=2004

em cảm ơn

a)

S1 = 2 - 4 + 6 - 8 + ..... + 1998 - 2000

    = (2 - 4) + (6 - 8) + ..... + (1998 - 2000)

    = (-2) + (-2) + ..... + (-2)

    = (-2).500

    = -1000

b)

S2 = 2 - 4 - 6 + 8 + ..... + 1994 - 1996 - 1998 + 2000

    = (2-4-6+8)+ ..... + (1994-1996-1998+2000)

    = 0 + 0 + ...... + 0

    = 0

  (2-4-6+8)+(10-12-14+16)+...+(1994-1996-1998+2000)

= (2+8-4-6)+(10+16-12-14)+...+(1994+2000-1996-1998)

= 0+0+....+0

= 0

-2000 nha bạn

Đặt: A=2-4-6+8+10-12-...+1994-1996-1998+2000

Ta có: A=2-4-6+8+10-12-...+1994-1996-1998+2000

 =>     A= (2 – 4 – 6 + 8) + (10- 12 – 14 + 16) + … + (1994 – 1996 – 1998 + 2000)

 =>     A = 0 + 0 + … + 0

 =>     A = 0

Vậy 2-4-6+8+10-12-...+1994-1996-1998+2000=0

a/ S1= 2-4+6-8+....+1998-2000

S1 = (2-4) + (6-8)+...+(1998-2000)

Số các  số là : (2000 - 2 ) : 2 + 1 = 1000 số

Số các cặp là: 1000 : 2 = 500 cặp

Mỗi cặp có giá trị là -2 nên : (-2) x 500 = -1000

Đáp số : âm 1000

olm duyệt

???

Ta có: 2-4+6-8+...+1998-2000

= (2-4)+(6-8)+...+(1998-2000)

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

= -2000

2-4-6+8+10-12-14+16+...+1994-1996-1998+2000

=( 2 - 4 - 6 + 8) + ( 10 - 12 - 14 + 16) + ................+ (1994 - 1996 - 1998 + 2000)

= 0 + 0 + ......... + 0

= 0