$\\\text{Quy đồng:}$ $\\\dfrac{9}{11}=\dfrac{90}{110}$ $\\\dfrac{10}{11}=\dfrac{100}{110}$ $\\\text{Gọi n là 2 phân số cần tìm}$ $\\\text{Ta có:$\dfrac{9}{11}<n<\dfrac{10}{11}$}$ $\\\text{Hay $\dfrac{90}{110}<n<\dfrac{100}{110}$}$ $\\\text{=>n ∈{$\dfrac{98}{110}$;$\dfrac{99}{110}$ }}$
