Câu hỏi của Trương Ngọc Linh

\(\dfrac{x+2020}{2}+\dfrac{2x+4040}{7}=\dfrac{x+2020}{5}+\dfrac{x+2020}{6}\)\(\Rightarrow\dfrac{x+2020}{2}+\dfrac{2x+4040}{7}-\dfrac{x+2020}{5}-\dfrac{x+2020}{6}=0\)\(\Rightarrow\dfrac{x+2020}{2}+\dfrac{2\left(x+2020\right)}{7}-\dfrac{x+2020}{5}-\dfrac{x+2020}{6}=0\)\(\Rightarrow\left(x+2020\right)\cdot\left(\dfrac{1}{2}+\dfrac{2}{7}-\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)Mà: \(\dfrac{1}{2}+\dfrac{2}{7}-\dfrac{1}{5}-\dfrac{1}{6}\ne0\)\(\Rightarrow x+2020=0\)\(\Rightarrow x=-2020\)Câu trả lời: \(\dfrac{x+2020}{2}+\dfrac{2x+4040}{7}=\dfrac{x+2020}{5}+\dfrac{x+2020}{6}\)\(\Rightarrow\dfrac{x+2020}{2}+\dfrac{2x+4040}{7}-\dfrac{x+2020}{5}-\dfrac{x+2020}{6}=0\)\(\Rightarrow\dfrac{x+2020}{2}+\dfrac{2\left(x+2020\right)}{7}-\dfrac{x+2020}{5}-\dfrac{x+2020}{6}=0\)\(\Rightarrow\left(x+2020\right)\cdot\left(\dfrac{1}{2}+\dfrac{2}{7}-\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)Mà: \(\dfrac{1}{2}+\dfrac{2}{7}-\dfrac{1}{5}-\dfrac{1}{6}\ne0\)\(\Rightarrow x+2020=0\)\(\Rightarrow x=-2020\)