Question

BT1: Tìm x, biết:

2) \(\dfrac{x+2017}{x+2018}=\dfrac{2020}{2021}\)

Answer 1

\(\dfrac{x+2017}{x+2018}=\dfrac{2020}{2021}\)

\(\Leftrightarrow1-\dfrac{x+2017}{x+2018}=1-\dfrac{2020}{2021}\)

\(\Leftrightarrow\dfrac{x+2018}{x+2018}-\dfrac{x+2017}{x+2018}=\dfrac{2021}{2021}-\dfrac{2020}{2021}\)

\(\Leftrightarrow\dfrac{\left(x+2018\right)-\left(x+2017\right)}{x+2018}=\dfrac{2021-2020}{2021}\)

\(\Leftrightarrow\dfrac{x+2018-x-2017}{x+2018}=\dfrac{1}{2021}\)

\(\Leftrightarrow\dfrac{\left(2018-2017\right)+\left(x+x\right)}{x+2018}=\dfrac{1}{2021}\)

\(\Leftrightarrow\dfrac{1}{x+2018}=\dfrac{1}{2021}\)

\(\Leftrightarrow x+2018=2021\)

\(\Leftrightarrow x=3\left(tm\right)\)

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