CẦU XIN CÁC BẠN

so sánh M với 3

M = 2019/2020 + 2020/2021 + 2021/2019

GIÚP MÌNH VỚI MỌI NGƯỜI ƠI

GẤP GẤP...... máy bay

M= 2019/2020 + 2020/2021 + 2021/2019

M = 1 

Mà 1 < 3

=> M < 3

#Ninh Nguyễn

M = 2019 / 2020 + 2020 / 2021 + 2021/ 2019

    = ( 1 - 1/2019 ) + ( 1 - 1/2020 ) + ( 1+ 2/2021 )

    = 3 - 1/2019 - 1/2020 + 2/2021

A - 3 = - 1/2019 - 1/2020 + 2/2021 > 0 

Vậy M > 3 

Ko bít có làm đúng ko nx nhưng bn cứ tham khảo bài mk nhé !

CẢM  ƠN X 100000000000000000

Ta có M=2019/2020+2020/2021+2021/2019

=>M=(1-1/2020)+(1-1/2021)+(1+2/2019)

=(1+1+1)+(2/2019-1/2020-1/2021)

=3+(1/2019+1/2019-1/2020-1/2021)

=3+(1/2019-1/2020)+(1/2019-1/2021)>3

Do 1/2019-1/2020>0

và 1/2019-1/2021>0

=>B>3

Vậy B>3

k cho mk nha

hok tốt=)))

m > 3 nhé

mong bạn giải rõ ra hộ mình

mình cảm ơn

\(m=\frac{2019}{2020}+\frac{2020}{2021}+\frac{2021}{2019}=1-\frac{1}{2020}+1-\frac{1}{2021}+1+\frac{2}{2019}\)

\(=3+\left(\frac{1}{2019}-\frac{1}{2020}\right)+\left(\frac{1}{2019}-\frac{1}{2021}\right)>3+0+0=3\)

mong mọi người giải rõ ra hộ mình với ạ

Giải:

Ta có: N=2019+2020/2020+2021

=>N=2019/2020+2021 + 2020/2020+2021

Vì 2019/2020 > 2019/2020+2021 ; 2020/2021 > 2020/2020+2021

=>M>N

Vậy ...

Chúc bạn học tốt!

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

            \(\dfrac{2020}{2021}>\dfrac{2020}{2020+2021}\)

\(\Rightarrow\dfrac{2019}{2020}+\dfrac{2020}{2021}>\dfrac{2019+2020}{2020+2021}\)

\(\Rightarrow M>N\)

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

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

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

Tham khảo:

 ta có: M=10^2020 +1 / 10^2019 +1

=> M/10= 10^2020 +1 / 10( 10^2019 +1 )

= 10^2020+1/ 10^2020 +10

=>  10/A=  10^2020 +10/10^2020 +1

=(10^2020 +1) +9/ 10^2020+1

=10^2020+1 /10^2020+1 + 9/10^2020+1

=1+ 9/10^2020+1

ta lại có: N=10^2021 +1/10^2020 +1

=> N/10= 10^2021+1/ 10(10^2020+1)

= 10^2021+1 / 10^2021+10

=> 10/N=10^2021+10 / 10^2021+1

=(10^2021+1) +9/10^2021+1

=10^2021+1/10^2021+1 +9/10^2021+1

=1+ 9/10^2021+1

ta thấy: 10/M>10N

=>M<N

\(M=\dfrac{10^{2020}+1}{10^{2019}+1}=1-\dfrac{9}{10^{2019}+1}\)

\(N=\dfrac{10^{2021}+1}{10^{2020}+1}=1-\dfrac{9}{10^{2020}+1}\)

Ta có: \(10^{2019}+1< 10^{2020}+1\)

\(\Leftrightarrow\dfrac{9}{10^{2019}+1}>\dfrac{9}{10^{2020}+1}\)

\(\Leftrightarrow-\dfrac{9}{10^{2019}+1}< -\dfrac{9}{10^{2020}+1}\)

\(\Leftrightarrow M< N\)

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