Giải phương trình $\frac{2019-x}{9}$+$\frac{19-x}{2000}$+$\frac{2009-x}{10}$+$\frac{4x}{2019}$=1 21/11/2021 Bởi Hadley Giải phương trình $\frac{2019-x}{9}$+$\frac{19-x}{2000}$+$\frac{2009-x}{10}$+$\frac{4x}{2019}$=1
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 $S = \left\{ 2019 + \dfrac{1}{\dfrac{1}{9} + \dfrac{1}{2000} + \dfrac{1}{10} – \dfrac{4}{2019}} \right\}$. Bình luận
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 $S = \left\{ 2019 + \dfrac{1}{\dfrac{1}{9} + \dfrac{1}{2000} + \dfrac{1}{10} – \dfrac{4}{2019}} \right\}$.