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

là 4000000 đó nha

a, Ta có : \(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1998}+\frac{x-4}{2000}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2000}{4}+\frac{x-1998}{6}+\frac{x-1996}{8}+\frac{x-1994}{10}\)

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

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

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

=> \(\left(x-2004\right)\left(\frac{1}{1994}+\frac{1}{1996}+\frac{1}{1998}+\frac{1}{2000}+\frac{1}{2002}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{10}=0\right)\)

=> \(x-2004=0\)

=> \(x=2004\)

Vậy phương trình có nghiệm là x = 2004 .

b, Ta có : \(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)

=> \(\frac{x-85}{15}-1+\frac{x-74}{13}-2+\frac{x-67}{11}-3+\frac{x-64}{9}-4=10-1-2-3-4=0\)

=> \(\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)

=> \(\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)

=> \(x-100=0\)

=> \(x=100\)

Vậy phương trình có nghiệm là x = 100 .

a/  (X+1)/35+1+(x+3)/33+1 =(x+5)/31+(x+7)/29+1+1

=>(x+36)/35+(x+36)/33-(x+36)/31-(x+36)/27=0

=>(X+36)(1/35+1/33-1/31-1/29)=0

=> x+36=0(vì c=vế 2 luôn luôn khác 0)

=>x=-36

b/ CMTT câu a 

trừ tung phân số cho 1 ta được x=2004

Ngu người khi ko biết làm bài lày