ko biết

bạn nào làm được thì giúp mình với còn bài này thì mình không biết làm. sorry nha

(2019-2018)^2020 và (2018-2017)^2019

=1^2020 và 1^2019

=1 và 1

Vì: 1=1

Nên: (2019-2018)^2020=(2018-2017)^2019

AI NÓI TỚ NÓI SAI, CÓ NÓI VỀ BÀI ĐÂU MÀ SAI ĐIÊN À MẤY BẠN KIA

\(x=\frac{2019^{2020}+1}{2019^{2019}+1}>\frac{2019^{2020}+1+2018}{2019^{2019}+1+2018}=\frac{2019^{2020}+2019}{2019^{2019}+2019}=\frac{2019\left(2019^{2019}+1\right)}{2019\left(2019^{2018}+1\right)}=\frac{2019^{2019}+1}{2019^{2018}+1}\)(1)

\(y=\frac{2019^{2019}+2020}{2019^{2018}+2020}< \frac{2019^{2019}+2020-2019}{2019^{2018}+2020-2019}=\frac{2019^{2019}+1}{2019^{2018}+1}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow x>y\)

Lời giải:

Ta có: 

\(A+1=\frac{2019^{2019}+2019^{2020}}{2019^{2019}-1}=\frac{2019^{2019}.2020}{2019^{2019}-1}\)

\(B+1=\frac{2019^{2019}+2019^{2018}}{2019^{2018}-1}=\frac{2019^{2018}.2020}{2019^{2018}-1}\) \(=\frac{2019^{2019}.2020}{2019^{2019}-2019}>\frac{2019^{2019}.2020}{2019^{2019}-1}\)

$\Rightarrow B+1>A+1$

$\Rightarrow B>A$

\(1-\frac{2018}{2019}=\frac{1}{2019}.\)

\(1-\frac{2019}{2020}=\frac{1}{2020}.\)

Ta có: 2019<2020 <=> \(\frac{1}{2019}>\frac{1}{2020}.\)

\(\Rightarrow-\frac{1}{2019}< -\frac{1}{2020}.\)

\(\Rightarrow1-\frac{1}{2019}< 1-\frac{1}{2020}.\)

\(\Rightarrow\frac{2018}{2019}< \frac{2019}{2020}.\)

2018/2019 < 2019/2020

để bằng 1 thì 2019/2020 phải cộng với 1/2020

để bằng 1 thì 2018/2019 phải cộng với 1/2019

vì 1/2020<1/2019 

=> 2019/2020>2018/2019

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

\(\frac{2018}{2019}=1-\frac{1}{2019}\)

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

hay \(\frac{2018}{2019}< \frac{2019}{2020}\)

Xét 2017 /2018 và 2018/2019

1-2017/2018=1/2018

1-2018/2019=1/2019

mà 1/2018>1/2019=>2017/2018<2018/2019

Tương tự có:2020/2019>2021/2020

=>2017/2018+2010/2019<2018/2019+2021/2020

Ta có:\(\frac{2018}{2019}\)<1\(\Rightarrow\)\(\frac{2018}{2019}\)>\(\frac{2018}{2019+2020}\)

           \(\frac{2019}{2020}\)<1\(\Rightarrow\)\(\frac{2019}{2020}\)>\(\frac{2019}{2019+2020}\)

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

\(\Rightarrow\)A>B

Vậy A>B

Ta có :\(A=\frac{2018}{2019}+\frac{2019}{2020}\)

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

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

Ta thấy :

\(\frac{2018}{2019}>\frac{2018}{2019+2020}\left(2019< 2019+2020\right)\)

\(\frac{2019}{2020}>\frac{2019}{2019+2020}\left(2020< 2019+2020\right)\)

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

Vậy \(A>B\)

~ Thiên Mã ~

2018/2019 + 2019/2020 và
(2018 + 2019)/(2019 + 2020) = 2018 /(2019 + 2020)+ 2018)/(2018 + 2019)
do 2018/2019>2018 /(2019 + 2020)
2018/2019 >2018)/(2019 + 2020)
nên
2018/2019 + 2019/2020 > (2018 + 2019)/(2019 + 2020)