Question

Tìm x biết:

\(\frac{x+1}{2019}+\)\(\frac{x+2}{2018}\)\(=\frac{x+3}{2017}+\)\(\frac{x+4}{2016}\)

Answer 1

\(\frac{x+1}{2019}+\frac{x+2}{2018}=\frac{x+3}{2017}+\frac{x+4}{2016}\)

\(\Leftrightarrow\left(\frac{x+1}{2019}-1\right)+\left(\frac{x+2}{2018}-1\right)=\left(\frac{x+3}{2017}-1\right)+\left(\frac{x+4}{2016}-1\right)\)

\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}=\frac{x+2020}{2017}+\frac{x+2020}{2016}\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\)

\(\Leftrightarrow x+2020=0:\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)\)

\(\Leftrightarrow x+2020=0\)

Còn lại tự làm :V

Answer 2

Lộn chỗ này , thay chút nha ! 

\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)=\left(\frac{x+3}{2017}+1\right)+\left(\frac{x+4}{2016}+1\right)\)

Sorry =))

Answer 3

                                                            Bài giải

Ta có : 

\(\frac{x+1}{2019}+\frac{x+2}{2018}=\frac{x+3}{2017}+\frac{x+4}{2016}\)

\(\frac{x+1}{2019}+1+\frac{x+2}{2018}+1=\frac{x+3}{2017}+1+\frac{x+4}{2016}+1\)

\(\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)=\left(\frac{x+3}{2017}+1\right)+\left(\frac{x+4}{2016}+1\right)\)

\(\frac{x+2020}{2019}+\frac{x+2020}{2018}=\frac{x+2020}{2017}+\frac{x+2020}{2016}\)

\(\frac{x+2020}{2019}+\frac{x+2020}{2018}-\frac{x+2020}{2017}-\frac{x+2020}{2016}=0\)

\(\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\)

Mà \(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\ne0\text{ }\)

          \(\Rightarrow\text{ }x+2020=0\)

          \(\Rightarrow\text{ }x=-2020\)          

Answer 4

\(\frac{x+1}{2019}+\frac{x+2}{2018}=\frac{x+3}{2017}+\frac{x+4}{2016}=0\)

\(\Leftrightarrow\frac{x+1}{2019}+\frac{x+2}{2018}-\frac{x+3}{2017}-\frac{x+4}{2016}=0\)

\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)-\left(\frac{x+3}{2017}+1\right)-\left(\frac{x+4}{2016}+1\right)=0\)

\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}-\frac{x+2020}{2017}-\frac{x+2020}{2016}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\)

Vì \(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\ne0\)\(\Rightarrow x+2020=0\)\(\Leftrightarrow x=-2020\)

Vậy \(x=-2020\)