giúp minkkkkkkkkkkkkkkkkkk

Bài 1:

$a)$

Ta có: $\frac{2021}{2022}=\frac{2022-1}{2022}=1-\frac{1}{2022}$

          $\frac{2020}{2021}=1-\frac{1}{2021}$

Đồng thời vì $ 2022>2021$ nên $\frac{-1}{2022}>\frac{-1}{2021}$

Vậy $\frac{2021}{2022}>\frac{2020}{2021}$

$b)$

Ta có: $\frac{2007}{2009}=1-\frac{2}{2009};\frac{2005}{2007}=1-\frac{2}{2007}$

Do đó $\frac{2007}{2009}>\frac{2005}{2007}$

$c)$

Ta có: $\frac{2021}{2023}=1-\frac{2}{2023};\frac{2017}{2019}=1-\frac{2}{2019}$

Do đó $\frac{2021}{2023}>\frac{2017}{2019}$

$d)$

Ta có:$\frac{2015}{2017}=1-\frac{2}{2017};\frac{2019}{2021}=1-\frac{2}{2021}$

Do đó $\frac{2015}{2017}<\frac{2019}{2021}$

Bài 2:

$a)$

Ta có: $\frac{2015}{2019}=1-\frac{4}{2019};\frac{2017}{2020}=1-\frac{3}{2020}$

Lại có:$-\frac{3}{2019}<-\frac{3}{2020}$

$\Rightarrow -\frac{3}{2019}-\frac{1}{2019}<-\frac{3}{2020}$

$\Rightarrow -\frac{4}{2019}<-\frac{3}{2020}$

Do đó: $\frac{2017}{2020}>\frac{2015}{2019}$

$b)$

Ta có: $\frac{1997}{1999}=1-\frac{2}{1999};\frac{1995}{1998}=1-\frac{3}{1998}$

Lại có:$-\frac{2}{1999}>-\frac{2}{1998}$

$\Rightarrow -\frac{2}{1998}-\frac{1}{1998}<-\frac{2}{1999}$

$\Rightarrow -\frac{3}{1998}<-\frac{2}{1999}$

Do đó: $\frac{1997}{1999}>\frac{1995}{1998}$

$c)$

Ta có: $\frac{2014}{2015}=1-\frac{1}{2015};\frac{2011}{2014}=1-\frac{3}{2014}$

Tương tự như những câu trên: $-\frac{1}{2015}>-\frac{3}{2014}

Do đó: $\frac{2014}{2015}>\frac{2011}{2014}$

$d)$

Tương tự như những câu trên ta dễ dàng nhận thấy:

$\frac{1768}{1771}<\frac{1775}{1777}$