Ta có: $12,(1)=12+0,(1)=12+\dfrac{1}{9}=\dfrac{109}{9}$ và $2,3(6)=2,3+\dfrac{1}{10}.0,(6)=2,3+\dfrac{1}{10}.6.0,(1)=2,3+\dfrac{1}{10}.6.\dfrac{1}{9}=\dfrac{71}{30}$ và $4,(21)=4+21.0,(01)=4+21.\dfrac{1}{99}=\dfrac{139}{33}$ ⇒ $[12,(1)-2,3(6)]:4,(21)=(\dfrac{109}{9}-\dfrac{71}{30}):\dfrac{139}{33}=\dfrac{9647}{4170}$ Chúc bạn học tốt !!!