Đáp án:
$$a) \dfrac{367}{120}\\ b)\dfrac{1998}{2001}<\dfrac{2013}{2016}<\dfrac{2020}{2023}.$$
Giải thích các bước giải:
$$a) 3\dfrac{2}{5}-\dfrac{2}{5} \times 2+\dfrac{11}{24}\\ =\dfrac{17}{5}-\dfrac{4}{5}+\dfrac{11}{24}\\ =\dfrac{13}{5}+\dfrac{11}{24}\\ =\dfrac{312}{120}+\dfrac{55}{120}\\ =\dfrac{367}{120}\\ b) \dfrac{2013}{2016}=1-\dfrac{3}{2016}\\ \dfrac{2020}{2023}=1-\dfrac{3}{2023}\\ \dfrac{1998}{2001}=1-\dfrac{3}{2001}$$
Ta có $$\dfrac{3}{2001}>\dfrac{3}{2016}>\dfrac{3}{2023}$$
$$\Rightarrow -\dfrac{3}{2001}<-\dfrac{3}{2016}<-\dfrac{3}{2023}$$
$$\Rightarrow 1 -\dfrac{3}{2001}<1-\dfrac{3}{2016}<1-\dfrac{3}{2023}$$
Hay $$\dfrac{1998}{2001}<\dfrac{2013}{2016}<\dfrac{2020}{2023}$$
Vậy $$\dfrac{1998}{2001}<\dfrac{2013}{2016}<\dfrac{2020}{2023}.$$
