\(\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}\\ ⇔\dfrac{x-23}{24}+\dfrac{x-3}{25}-\dfrac{x-23}{26}-\dfrac{x-23}{27}=0\\ ⇔(x-23)\bigg(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\bigg)=0\\ ⇔x-23=0\ \text{$\bigg($Vì $\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\neq 0\bigg)$}\\ ⇔x=23\\ \text{Vậy phương trình có tập nghiệm S = $\{23\}$}\)
