Question

Tìm x

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

Answer 1

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

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

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

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

\(\Rightarrow x+1=0\)

\(x=-1\)

Answer 2

\(\Leftrightarrow\frac{x+1}{2015}+\frac{x+1}{2016}-\frac{x+1}{2017}-\frac{x+1}{2018}=0\)

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

\(\Leftrightarrow x+1=0\) ( vì \(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\ne0\))

\(\Leftrightarrow x=-1\)

Answer 3

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

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

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

=>x+1=0=>x=-1

Vậy x=-1

Answer 4

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

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

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

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

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

Vì \(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\ne0\)=> x+1=0

=>x=0-1

=>x=-1

Answer 5

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

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

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

=> x + 1 = 0

=> x = -1