Đáp án:
$S = \left\{ 2019 + \dfrac{1}{\dfrac{1}{9} + \dfrac{1}{2000} + \dfrac{1}{10} - \dfrac{4}{2019}} \right\}$.
Giải thích các bước giải:
Ptrinh đã cho tương đương vs
$\dfrac{2019-x}{9} + \dfrac{19-x}{2000} + 1 + \dfrac{2009-x}{10} + 1 + \dfrac{4x}{2019} - 4 = -1$
$\Leftrightarrow \dfrac{2019-x}{9} + \dfrac{2019-x}{2000} + \dfrac{2019 - x}{10} + \dfrac{4x - 4.2019}{2019} = -1$
$\Leftrightarrow (2019-x) \left( \dfrac{1}{9} + \dfrac{1}{2000} + \dfrac{1}{10} - \dfrac{4}{2019} \right) = -1$
$\Leftrightarrow x-2019 = \dfrac{1}{\dfrac{1}{9} + \dfrac{1}{2000} + \dfrac{1}{10} - \dfrac{4}{2019}}$
$\Leftrightarrow x = 2019 + \dfrac{1}{\dfrac{1}{9} + \dfrac{1}{2000} + \dfrac{1}{10} - \dfrac{4}{2019}}$
Vậy $S = \left\{ 2019 + \dfrac{1}{\dfrac{1}{9} + \dfrac{1}{2000} + \dfrac{1}{10} - \dfrac{4}{2019}} \right\}$.