Question

Tìm nghiêm là số tự nhiên của phương trình : 

X²-2022x-1/2022+ x²-2022x-2/2021= x²-2022x-4/2019+x²-2022x-5/ 2018

 

Answer 1

\(\dfrac{x^2-2022x-1}{2022}+\dfrac{x^2-2022x-2}{2021}=\dfrac{x^2-2022x-4}{2019}+\dfrac{x^2-2022x-5}{2018}\)

=>\(\left(\dfrac{x^2-2022x-1}{2022}-1\right)+\left(\dfrac{x^2-2022x-2}{2021}-1\right)=\left(\dfrac{x^2-2022x-4}{2019}-1\right)+\left(\dfrac{x^2-2022x-5}{2018}-1\right)\)

=>\(\dfrac{x^2-2022x-2023}{2022}+\dfrac{x^2-2022x-2023}{2021}-\dfrac{x^2-2022x-2023}{2019}-\dfrac{x^2-2022x-2023}{2018}=0\)

=>\(x^2-2022x-2023=0\)

=>(x-2023)(x+1)=0

=>\(\left[{}\begin{matrix}x=2023\\x=-1\end{matrix}\right.\)