Question

Giải giúp mình với ạ:)

Answer 1

\(\Leftrightarrow\left(\dfrac{x^2+99x-1}{99}-1\right)+\left(\dfrac{x^2+99x-2}{98}-1\right)+\left(\dfrac{x^2+99x-3}{97}-1\right)=\left(\dfrac{x^2+99x-4}{96}-1\right)+\left(\dfrac{x^2+99x-5}{95}-1\right)+\left(\dfrac{x^2+99x-6}{94}-1\right)\)

=>x^2+99x-100=0

=>x=-100 hoặc x=1