Trả lời

So sánh cái nào vs cái nào ạ

sao chỉ thấy có 1 vế ạ !

vi 2018/2019<1

   2019/2020<1

   2020/2021<1

nen 2018/2019 + 2019/2020 + 2020/2021<1+1+1=3

                                         Bài giải

                Ta có : 

\(\frac{2018}{2019}< 1\)

\(\frac{2019}{2020}< 1\)

\(\frac{2020}{2021}< 1\)

\(\Rightarrow\text{ }\frac{2018}{2019}+\frac{2019}{2020}+\frac{2020}{2021}< 1+1+1=3\)

\(\text{Vậy }\frac{2018}{2019}+\frac{2019}{2020}+\frac{2020}{2021}< 3\)

Đặt \(A=\frac{\frac{1}{2020}+\frac{2}{2019}+\frac{3}{2018}+...+\frac{2019}{2}+\frac{2020}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)

\(A=\frac{1+\left(\frac{1}{2020}+1\right)+\left(\frac{2}{2019}+1\right)+\left(\frac{3}{2018}+1\right)+...+\left(\frac{2019}{2}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)

\(A=\frac{\frac{2021}{2021}+\frac{2021}{2020}+\frac{2021}{2019}+...+\frac{2021}{2}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)

\(A=\frac{2021\left(\frac{1}{2021}+\frac{1}{2020}+\frac{1}{2019}+...+\frac{1}{2}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}=2021\)

Ta có: \(\frac{2019}{2020}>\frac{2019}{2020+2021};\frac{2020}{2021}>\frac{2020}{2020+2021}\)

=> \(\frac{2019}{2020}+\frac{2020}{2021}>\frac{2019}{2020+2021}+\frac{2020}{2020+2021}=\frac{2019+2020}{2020+2021}\)

=> A > B.

Nhân 79 vào cả hai vế rồi tự làm

LÀm đỡ mk tí mk ko có nhiều tgian vi còn 5 đề nữa

Lời giải:
$A=1-\frac{1}{2019}+1-\frac{1}{2020}+1-\frac{1}{2021}+1+\frac{3}{2018}$

$=4+(\frac{1}{2018}-\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2020}+\frac{1}{2018}-\frac{1}{2021})$

$> 4+0+0+0+0=4$

a) Ta có : \(\frac{-60}{12}=-5=-\frac{25}{5}\)

\(-0,8=-\frac{8}{10}=-\frac{4}{5}\)

Mà -25 < -4 nên \(\frac{-25}{5}< \frac{-4}{5}\)=> \(\frac{-60}{12}< -0,8\)

b) Ta có : \(\frac{2020}{2019}=1+\frac{1}{2019}\)

\(\frac{2021}{2020}=1+\frac{1}{2020}\)

Vì \(\frac{1}{2019}>\frac{1}{2020}\)nên \(\frac{2020}{2019}>\frac{2021}{2020}\)

c) \(\frac{10^{2018}+1}{10^{2019}+1}=\frac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\frac{10^{2019}+10}{10^{2019}+1}=\frac{10^{2019}+1+9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)(1)

\(\frac{10^{2019}+1}{10^{2020}+1}=\frac{10\left(10^{2019}+1\right)}{10^{2020}+1}=\frac{10^{2020}+10}{10^{2020}+1}=\frac{10^{2020}+1+9}{10^{2020}+1}=1+\frac{9}{10^{2020}+1}\)(2)

Đến đây tự so sánh rồi nhé

N =2019+2020/2020+2021

=2019/2020+2021  +   2020/2020+2021

Ta có:

2019/2020>2019/2020+2021

2020/2021 > 2020/2020+2021

=>M>N

b,

          \(B=\frac{2018+2019}{2019+2020}=\frac{2018}{2019+2020}+\frac{2019}{2019+2020}\)

  Ta thấy :

\(\frac{2018}{2019}>\frac{2018}{2019+2020}\)

\(\frac{2019}{2020}>\frac{2019}{2019+2020}\)

Từ đó , suy ra :

\(\frac{2018}{2019}+\frac{2019}{2020}>\frac{2018+2019}{2019+2020}\)

                                       Vậy...

                                                        #Louis

Ta có :

\(\frac{205}{321}< \frac{205}{315}\)

\(\frac{214}{315}>\frac{205}{315}\)

\(\Leftrightarrow\frac{205}{321}< \frac{214}{315}\)

bn forever alone giai thich giup mk vs.tai sao lai the