Câu hỏi của Khánh Linh

Question

BT1: Tìm x $\in$ Z biết:

1) $\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}$

Nguyễn Thanh Hằng · Accepted Answer

$\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}$

$\Leftrightarrow\left(\dfrac{1-x}{2017}+1\right)+\left(\dfrac{2-x}{2016}+1\right)=\left(\dfrac{3-x}{2015}+1\right)+\left(\dfrac{4-x}{2014}+1\right)$

$\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}=\dfrac{2018-x}{2015}+\dfrac{2018-x}{2014}$

$\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}-\dfrac{2018-x}{2015}-\dfrac{2018-x}{2014}=0$

$\Leftrightarrow\left(2018-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0$

Mà $\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0$

$\Leftrightarrow2018-x=0\Leftrightarrow x=2018$