\(-x-\frac{9}{2004}=-\frac{1}{2003}\)

\(\Rightarrow-x=\frac{-1}{2003}+\frac{9}{2004}\)

\(\Rightarrow-x=\frac{-1}{2003}+\frac{9}{2004}\)

\(\frac{5}{9}-x=1-2004\)

\(\Rightarrow\frac{5}{9}-x=-2003\)

\(\Rightarrow x=\frac{5}{9}-\left(-2003\right)\)

\(\Rightarrow x=\frac{18032}{9}\)

a )

\(-x-\frac{9}{2004}=-\frac{1}{2003}\)

\(-x=-\frac{1}{2003}+\frac{9}{2004}\)

Số lớn quá 

b ) \(\frac{5}{9}-x=\frac{1}{2004}\)

\(x=\frac{5}{9}-\frac{1}{2004}\)

\(x=\frac{3337}{6012}\)

\(\frac{2006\times2004-9}{1995+2004\times2005}=\frac{\left(2005+1\right)\times2004-9}{1995+2004\times2005}\)

\(=\frac{2005\times2004+2004-9}{2004\times2005+1995}\)

\(=\frac{2005\times2004+1995}{1995+2004\times2005}\)

\(=1\)

2006x 2004 -9/1995 +2004x 2005

=(2005+1)x2004 -9/1995+2004x2005

=2005x2004+2004 x1-9/1995+2005x2004

=2005x2004+2004-9/1995+2004x2005

=2005x2004+1995/1995+2005x2004

=1

2006x 2004-9/ 1995+2004x 2005

=(2005+1)x 2004- 9/ 1995+2004x2005

= 2005x2004+2004-9(nhân phân phối)/ 1995+2004x2005

=2005x2004+1995/ 1995+2004x2005

=1(cả mẫu lẫn tử đều giống nhau nên thu gọn)

click cho tớ nhé

\(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{1999+2004}\).

Có sai đề không vậy???

Sửa đề một chút :v

\(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{1999\cdot2004}\)

\(=\frac{1}{5}\left[\frac{5}{4\cdot9}+\frac{5}{9\cdot14}+\frac{5}{14\cdot19}+...+\frac{5}{1999\cdot2004}\right]\)

\(=\frac{1}{5}\left[\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{1999}-\frac{1}{2004}\right]\)

\(=\frac{1}{5}\left[\frac{1}{4}-\frac{1}{2004}\right]\)

\(=\frac{1}{5}\cdot\frac{125}{501}=\frac{25}{501}\)

Đặt A =1/4 x 9 + 1/9 x 14 + 1/14 x 19 +...+ 1/1999 + 2004. Ta có:

        A= 1/4 x 9 + 1/9 x 14 + 1/14 x 19 +...+ 1/1999 + 2004

      5A= 5/4 x 9 + 5/9 x 14 + 5/14 x 19 +...+ 5/1999 + 2004

      5A= 1/4 - 1/9 + 1/9 - 1/14 + 1/14 - 1/19 +...+ 1/1999 - 1/2004

      5A= 1/4 - 1/2004

        A= (1/4 - 1/2004)/5

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