Giải phương trình : 4-x/2018-2=3-x/2019-x/1011

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1. Đáp án:

   `S={2022}.`

   Giải thích các bước giải:

   `(4-x)/2018-2=(3-x)/2019-x/1011`

   `(4-x)/2018-2+3 =(3-x)/2019-x/1011 +3 `

   `<=>(4-x)/2018+1=(3-x)/2019+1 -x/1011 +2`

   `<=>(2022-x)/2018 = (2022-x)/2019 + (2022-x)/2011`

   `<=>(2022-x)/2018 – (2022-x)/2019 – (2022-x)/2011=0`

   `<=>(2022-x)(1/2018 -1/2019 -1/2011)=0`

   `<=>2022-x=0( vì 1/2018 -1/2019 -1/2011 \ne 0)`

   `<=>x=2022`

   Vậy phương trình có tập nghiệm `S={2022}.`

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2. Đáp án:

   $\begin{array}{l}\dfrac{4-x}{2018} – 2 = \dfrac{3-x}{2019} – \dfrac{x}{1011} \\ \Leftrightarrow \dfrac{4-x}{2018} – 2 + 3 = \dfrac{3-x}{2019} – \dfrac{x}{1011} + 3 \\ \Leftrightarrow \dfrac{4-x}{2018} + 1 = \dfrac{3-x}{2019} + 1 – \dfrac{x}{1011} + 2 \\ \Leftrightarrow \dfrac{4-x+2018}{2018} = \dfrac{3-x-2019}{2019} + \dfrac{-x+1011.2}{1011} \\ \Leftrightarrow \dfrac{2022-x}{2018} = \dfrac{2022-x}{2019} + \dfrac{2022-x}{1011} \\ \Leftrightarrow  \dfrac{2022-x}{2018} – \dfrac{2022-x}{2019} – \dfrac{2022-x}{1011} = 0 \\ \Leftrightarrow ( 2022-x ) . ( \dfrac{1}{2018} – \dfrac{1}{2019} – \dfrac{1}{1011} ) = 0 \\ \Leftrightarrow 2022-x=0 \ \ (\text{Vì} \ \dfrac{1}{2018} – \dfrac{1}{2019} – \dfrac{1}{1011} \ne 0 )\\ \Leftrightarrow x=2022 \\ \text{Vậy} \ S=\{2022\} \end{array}$

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