\(\dfrac{2004\times2005+2006\times6-6}{2005\times1997+4\times2005}\\ =\dfrac{2004\times2005+\left(2005+1\right)\times6-6}{2005\times\left(1997+4\right)}\\ =\dfrac{2004\times2005+2005\times6+6-6}{2005\times2001}\\ =\dfrac{2005\times\left(2004+6\right)}{2005\times2001}\\ =\dfrac{2010}{2001}\\ =\dfrac{670}{667}\)

Mình mới làm được phần (b) thôi bạn Natsu Dragneel , phần (a) khó quá !

\(b,\frac{167x198+98}{198x168-100}\)

\(=\frac{167x198+98}{198x\left(167+1\right)-100}\)

\(=\frac{167x198+98}{198x167+198x1-100}\)

\(=\frac{167x198+98}{198x167+98}=1\)

\(\frac{670}{667}\)

3958021

\(\frac{1999.2000+2001.5-5}{504.2004+500.2000}\)

Cài đề phải gì không 

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

\(\frac{1999x2001-1}{1998+1999x2000}x\frac{7}{5}=\frac{1999x\left(2000+1\right)-1}{1998+1999x2000}x\frac{7}{5}\)

\(=\frac{1999x2000+1999-1}{1998+1999x2000}x\frac{7}{5}\)

\(=\frac{1999x2000+1998}{1998+1999x2000}x\frac{7}{5}\)

\(=1x\frac{7}{5}\)

\(=\frac{7}{5}\)

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

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

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