Giải:
\(11\dfrac{3}{4}-\left(6\dfrac{5}{6}-4\dfrac{1}{2}\right)+1\dfrac{2}{3}\)
\(=\dfrac{47}{4}-\left(\dfrac{41}{6}-\dfrac{9}{2}\right)+\dfrac{5}{3}\)
\(=\dfrac{47}{4}-\dfrac{41}{6}+\dfrac{9}{2}+\dfrac{5}{3}\)
\(=\dfrac{141}{12}-\dfrac{82}{12}+\dfrac{54}{12}+\dfrac{20}{12}\)
\(=\dfrac{141-82+54+20}{12}\)
\(=\dfrac{133}{12}\)
Chúc bạn học tốt!