Question

[Phan Thế Phong]

BT: Tìm x biết:

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

Answer 1

[Vivian]

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

=> \(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)

=> (x+2018).\(\left(\dfrac{1}{2014}+\dfrac{1}{2015}\right)=\left(x+2018\right).\left(\dfrac{1}{2016}+\dfrac{1}{2017}\right)\)

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

Mà \(\dfrac{1}{2014}>0;\dfrac{1}{2015}>0;\dfrac{1}{2016}>0;\dfrac{1}{2017}>0\)

=>\(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\ne0\)

=> \(x+2018=0\)

=>x = 0-2018

=> x = 0+(-2018)

=> x = -2018