Question

\(\frac{x+4}{2015}+\frac{x+3}{2016}=\frac{x+2}{2017}+\frac{x+1}{2018}\text{ }\text{ }\)

Answer 1

\(\frac{x+4}{2015}+\frac{x+3}{2016}=\frac{x+2}{2017}+\frac{x+1}{2018}\)

\(\Rightarrow\frac{x+4}{2015}+1+\frac{x+3}{2016}+1=\frac{x+2}{2017}+1+\frac{x+1}{2018}+1\)

\(\Rightarrow\frac{x+4+2015}{2015}+\frac{x+3+2016}{2016}=\frac{x+2+2017}{2017}+\frac{x+1+2018}{2018}\)

\(\Rightarrow\frac{x+2019}{2015}+\frac{x+2019}{2016}-\frac{x+2019}{2017}-\frac{x+2019}{2018}=0\)

\(\Rightarrow\left(x+2019\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)

Vì \(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\ne0\)

=> x + 2019 = 0

=> x             = -2019

Vậy x = -2019

Answer 2

x=-2019

Answer 3

thank you MMS_Hồ Khánh Châu

Answer 4

\(\frac{x+4}{2015}+\frac{x+3}{2016}=\frac{x+2}{2017}+\frac{x+1}{2018}\)

=> \(\frac{x+4}{2015}+\frac{x+3}{2016}-\frac{x+2}{2017}-\frac{x+1}{2018}=0\)

=> \(\left(\frac{x+4}{2015}+1\right)+\left(\frac{x+3}{2016}+1\right)-\left(\frac{x+2}{2017}+1\right)-\left(\frac{x+1}{2018}+1\right)=0\)

=> \(\frac{x+2019}{2015}+\frac{x+2019}{2016}-\frac{x+2019}{2017}-\frac{x+2019}{2018}=0\)

=> \(\left(x+2019\right).\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\right)=0\)

=> \(x+2019=0\)hoặc \(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}=0\)

Mà \(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}\)khác 0

=> \(x+2019=0\)

=> \(x=0-2019\)

=> \(x=\left(-2019\right)\)