So Sánh M và N biết :
M= 10^2020+1/10^2019+1 và N=10^2021+1/10^2020+1
giải nhanh giúp mik với ạ =))
K
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L
28 tháng 1 2020
\(M=\frac{10^{2018}+1}{10^{2019}+1}\)
\(\Rightarrow10M=\frac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\frac{10^{2019}+1+9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)
\(N=\frac{10^{2019}+1}{10^{2020}+1}\)
\(\Rightarrow10N=\frac{10\left(10^{2019}+1\right)}{10^{2020}+1}=\frac{10^{2020}+1+9}{10^{2020}+1}=1+\frac{9}{10^{2020}+1}\)
Ta co: \(\frac{9}{10^{2019}+1}>\frac{9}{10^{2020}+1}\) ma \(1=1\)
\(\Rightarrow1+\frac{9}{10^{2019}+1}>1+\frac{9}{10^{2020}+1}\)
\(\Rightarrow10M>10N\)
\(\Rightarrow M>N\)
11 tháng 3 2022
ta có :
A = \(\dfrac{5^{2020}+1}{5^{2020}+1}\)
B = \(\dfrac{5^{2019}+1}{5^{2020}+1}\)
\(\Leftrightarrow\) B < A
VT
1
11 tháng 3 2021
Ta có \(b-a=9.10^{2019}-\dfrac{9}{10^{2021}}>0\Rightarrow b>a\).
XO
30 tháng 7 2020
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
VT
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
2 tháng 5 2023
A = \(\dfrac{5^{2020}+1}{5^{2021}+1}\) ⇒ A \(\times\) 10 = 2 \(\times\)5 \(\times\) \(\dfrac{5^{2020}+1}{5^{2021}+1}\) =2\(\times\) \(\dfrac{5^{2021}+5}{5^{2021}+1}\)
10A =2 \(\times\) \(\dfrac{5^{2021}+5}{5^{2021}+1}\) = 2 \(\times\)(1 + \(\dfrac{4}{5^{2021}+1}\) )= 2 + \(\dfrac{8}{5^{2021}+1}\) >2
B = \(\dfrac{10^{2019}+1}{10^{2020}+1}\) ⇒ B \(\times\) 10 = 10 \(\times\) \(\dfrac{10^{2019}+1}{10^{2020}+1}\)= \(\dfrac{10^{2020}+10}{10^{2020}+1}\)
10B = \(\dfrac{10^{2020}+10}{10^{2020}+1}\) = 1 + \(\dfrac{9}{10^{2020}+1}\) < 2
10A > 2 > 10B ⇒ 10A>10B ⇒ A>B
M
1
20 tháng 5 2021
Giải:
Ta có:
A=\(\dfrac{10^{2019}-1}{10^{2020}+1}\)
10A=\(\dfrac{10^{2020}-10}{10^{2020}+1}\)
10A=\(\dfrac{10^{2020}+1-11}{10^{2020}+1}\)
10A=\(1+\dfrac{-11}{10^{2020}+1}\)
Tương tự:
B=\(\dfrac{10^{2020}-1}{20^{2021}+1}\)
10B=\(1+\dfrac{-11}{10^{2021}+1}\)
Vì \(\dfrac{-11}{10^{2020}+1}< \dfrac{-11}{10^{2021}+1}\) nên 10A<10B
⇒A<B
Chúc bạn học tốt!
NT
0
16 tháng 5 2022
Ta có:
\(10A=\dfrac{10\left(10^{2020}+1\right)}{10^{2021}+1}=\dfrac{10^{2021}+10}{10^{2021}+1}=1+\dfrac{9}{10^{2021}+1}\)
\(10B=\dfrac{10\left(10^{2021}+1\right)}{10^{2022}+1}=\dfrac{10^{2022}+10}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)
⇒ \(10A>10B\) ( vì \(\dfrac{9}{10^{2021}+1}>\dfrac{9}{10^{2022}+1}\) )
Suy ra: \(A>B\)
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: \(10^{2019}+1< 10^{2020}+1\)
\(\Leftrightarrow\dfrac{9}{10^{2019}+1}>\dfrac{9}{10^{2020}+1}\)
\(\Leftrightarrow-\dfrac{9}{10^{2019}+1}< -\dfrac{9}{10^{2020}+1}\)
\(\Leftrightarrow M< N\)