Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(\frac{15}{106}\)và \(\frac{21}{133}\)
Ta có:
\(\frac{15}{106}< \frac{15}{100}=\frac{3}{20}=\frac{21}{140}< \frac{21}{133}\)
\(\Rightarrow\frac{15}{106}< \frac{21}{133}\)
Vậy ........
b, \(\frac{31}{100}\)và \(\frac{89}{150}\)
Ta có:
\(\frac{31}{100}< \frac{31}{93}=\frac{1}{3}=\frac{50}{150}< \frac{89}{150}\)
\(\Rightarrow\frac{31}{100}< \frac{89}{150}\)
Vậy........
c, \(\frac{2020}{2019}\)và \(\frac{2021}{2020}\)
Ta có:
\(\frac{2020}{2019}-1=\frac{1}{2019}\) ;
\(\frac{2021}{2020}-1=\frac{1}{2020}\)
Vì \(\frac{1}{2019}>\frac{1}{2020}\)
\(\Rightarrow\frac{2020}{2019}-1>\frac{2021}{2020}-1\)
\(\Rightarrow\frac{2020}{2019}>\frac{2021}{2020}\)
Vậy .........
d, n+2019/n+2021 và n+2020/n+2022
Câu d bn tự lm nhé
\(\dfrac{-11}{-32}>\dfrac{16}{49}\)
\(\dfrac{-2020}{-2021}>\dfrac{-2021}{2022}\)
Ta có : A = \(\frac{10^{2020}+1}{10^{2021}+1}\)
=> 10A = \(\frac{10^{2021}+10}{10^{2021}+1}=1+\frac{9}{10^{2021}+1}\)
Lại có : \(B=\frac{10^{2021}+1}{10^{2022}+1}\)
=> \(10B=\frac{10^{2022}+10}{10^{2022}+1}=1+\frac{9}{10^{2022}+1}\)
Vì \(\frac{9}{10^{2022}+1}< \frac{9}{10^{2021}+1}\)
=> \(1+\frac{9}{10^{2022}+1}< 1+\frac{9}{10^{2022}+1}\)
=> 10B < 10A
=> B < A
b) Ta có : \(\frac{2019}{2020+2021}< \frac{2019}{2020}\)
Lại có : \(\frac{2020}{2020+2021}< \frac{2020}{2021}\)
=> \(\frac{2019}{2020+2021}+\frac{2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)
=> \(\frac{2019+2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)
=> B < A
1) \(16^{2020}+\dfrac{1}{16^{2021}}+1\)
\(=16^{2021}\div16^{2020}+1\)
\(=16+1\)
\(=17\)
2) \(16^{2021}+\dfrac{1}{16^{2022}}+1\)
\(=16^{2022}\div16^{2021}+1\)
\(=16+1\)
= 17
Vì 17=17 nên \(16^{2020}+\dfrac{1}{16^{2021}}+1=16^{2021}+\dfrac{1}{16^{2022}}+1\)

1−6364=1641-6364=164
1−3233=1331-3233=133
Vì 164<133⇒6364>3233164<133⇒6364>3233
Vậy 6364>3233
20202019−1=1201920202019-1=12019
20222021−1=1202120222021-1=12021
Mà 12019>1202112019>12021
⇒20202019>20222021⇒20202019>20222021
Vậy 20202019>20222021
\(\dfrac{63}{64}\) = 1 - \(\dfrac{1}{64}\)
\(\dfrac{32}{33}\) = 1 - \(\dfrac{1}{33}\)
Vì \(\dfrac{1}{33}\) > \(\dfrac{1}{64}\)
Vậy \(\dfrac{63}{64}\) > \(\dfrac{32}{33}\)
\(\dfrac{2020}{2019}\)= 1 + \(\dfrac{1}{2019}\)
\(\dfrac{2022}{2021}\) = 1 + \(\dfrac{1}{2021}\)
Vì \(\dfrac{1}{2019}\) > \(\dfrac{1}{2021}\)
Vậy \(\dfrac{2020}{2019}\) > \(\dfrac{2022}{2021}\)
Lời giải:
$\frac{63}{64}=1-\frac{1}{64}> 1-\frac{1}{33}=\frac{32}{33}$
-------------------------
$\frac{2020}{2019}=1+\frac{1}{2019}> 1+\frac{1}{2021}=\frac{2022}{2021}$
cái này so sánh = phần bù / phần hơn
quy đồng 2019 với 2021 lâu lắm