Giải:

Ta có: 

2019.2020-1/2019.2020= 2019.2020/2019.2020 - 1/2019.2020

                                       =1-1/2019.2020

Tương tự:

2020.2021-1/2020.2021= 1-1/2020.2021

Vì 1/2019.2020 > 1/2020.2021 nên -1/2019.2020 < -1/2020.2021

(vì là số nguyên âm)

⇒ 1-1/2019.2020 < 1-1/2020.2021

⇔ 2019.2020-1/2019.2020 < 2020.2021-1/2020.2021

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hinh nhu sai de roi

\(A=1.2.3...100-1.2.3.4...99-1.2.3.4....99^2\)

\(=1.2.3....99.\left(100-1\right)-1.2.3...98.99^2\)\(=1.2.3...99^2-1.2.3...99^2=0\)

\(=1.2.....2014.\left(2015-1-2014\right)=1.2....2014.0=0\)

1.2.3.4.....9-1.2.3.4.....8-1.2.3.4.....8²

=(1.2.3.4.....8).(9-8)

=(1.2.3.4.....8).1

=1.2.3.4.....8

=40320

Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2019\cdot2020}+\dfrac{1}{2020\cdot2021}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2019}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2021}\)

\(=\dfrac{1}{1}-\dfrac{1}{2021}=\dfrac{2021}{2021}-\dfrac{1}{2021}\)

\(=\dfrac{2020}{2021}\)

mà \(\dfrac{2020}{2021}< \dfrac{2021}{2021}=1\)

nên A<1

Ta có\(A=1.2.3....2019-1.2.3...2018-1.2.3...2017.2018^2\)

            \(=1.2.3...2018.\left(2019-1-2018\right)\)

            \(=1.2.3....2018.0\)

            \(=0\)