`1)`
\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right)\cdot\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}+\frac{3}{35}\right)\cdot\left(\frac{-4}{3}\right)}\)
\(=\frac{\left(\frac{18}{60}-\frac{16}{60}-\frac{21}{60}\right)\cdot\frac{5}{19}}{\left(\frac{5}{70}+\frac{10}{70}+\frac{6}{70}\right)\cdot\left(\frac{-4}{3}\right)}\)
\(=\frac{\left(\frac{-19}{60}\right)\cdot\frac{5}{19}}{\frac{3}{10}\cdot\left(\frac{-4}{3}\right)}\)
\(=\frac{\frac{-1}{12}}{\frac{-2}{5}}\)
\(=\left(\frac{-1}{12}\right):\left(\frac{-2}{5}\right)\)
\(=\frac{5}{24}\)
`2)`
\(\dfrac{x+4}{2000}\) + \(\dfrac{x+3}{2001}\) =\(\dfrac{x+2}{2002}\) + \(\dfrac{x+1}{2003}\)
`<=>` \(\dfrac{x+4}{2000}\) + 1 + \(\dfrac{x+3}{2001}\) +1 = \(\dfrac{x+2}{2002}\) + 1 + \(\dfrac{x+1}{2003}\) `+ 1`
`<=>`\(\dfrac{x+4}{2000}\)+\(\dfrac{2000}{2000}\)+\(\dfrac{x+3}{2001}\) \(\dfrac{2001}{2001}\) = \(\dfrac{x+2}{2002}\)+\(\dfrac{2002}{2002}\)+\(\dfrac{x+1}{2003}\)+\(\dfrac{2003}{2003}\)
`<=>` \(\dfrac{x+4+2000}{2000}\)+\(\dfrac{x+3+2001}{2001}\) = \(\dfrac{x+2+2002}{2002}\)+ \(\dfrac{x+1+2003}{2003}\)
`<=>` \(\dfrac{x+2004}{2000}\) + \(\dfrac{x+2004}{2001}\) - \(\dfrac{x+2004}{2002}\) - \(\dfrac{x+2004}{2003}\) `= 0`
`<=> (x+2004)`(\(\dfrac{1}{2000}\) + \(\dfrac{1}{2001}\) - \(\dfrac{1}{2002}\) -\(\dfrac{1}{2003}\)) = 0
Mà \(\dfrac{1}{2000}\) + \(\dfrac{1}{2001}\) - \(\dfrac{1}{2002}\) - \(\dfrac{1}{2003}\) ` \ne 0`
Nên `x+2004=0`
`=>x=0-2004`
`=> x = -2004`
Vậy `S = -2004`
