Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
Tính tổng:
S1=1+(-2)+3+(-4)+...+2015+(-2016).
S1=[1+(-2)]+[3+(-4)]+...+[2015+(-2016)]
S1=-1+(-1)+...+(-1) ( có 1008 số -1 )
S1=-1.1008
S1=-1008
S3=1+(-3)+5+(-7)+...2013+(-2015)
S3=[1+(-3)]+[5+(-7)]+...+[2013+(-2015)]
S3=-2+(-2)+...+(-2) ( có 1008 số -2)
S3=-2016
ta có \(A=\frac{2020^{10}+2}{2020^{11}+2}=>2020A=\frac{2020^{11}+4040}{2020^{11}+2}=1+\frac{4038}{2020^{11}+2}\)(1)
\(B=\frac{2020^{11}+2}{2020^{12}+2}=>2020B=\frac{2020^{12}+4040}{2020^{12}+2}=1+\frac{4038}{2012^{12}+2}\)(2)
từ 1 and 2 => 2020B<2020A
=> A>B
Ta có B=\(\frac{2020^{11}+2}{2020^{12}+2}\)
suy ra \(B< \frac{\left(2020^{11}+2\right)+2018}{\left(2020^{12}+2\right)+2018}=\frac{2020^{11}+2020}{2020^{12}+2020}=\frac{2020\left(2020^{10}+2\right)}{2020\left(2020^{11}+2\right)}=\frac{2020^{10}+2}{2020^{11}+2}\)
nên A > B
\(10A=\dfrac{10^{2015}+2016+9\cdot2016}{10^{2015}+2016}=1+\dfrac{18144}{10^{2015}+2016}\)
\(10B=\dfrac{10^{2016}+9+18144}{10^{2016}+2016}=1+\dfrac{18144}{10^{2016}+2016}\)
mà \(\dfrac{18144}{10^{2015}+2016}>\dfrac{18144}{10^{2016}+2016}\)
nên A>B
Trả lời :
- 2 bn kia ở trong câu hỏi này có ai làm đúng đâu.
- Chúc bạn học tốt !
- Tk cho mk nha !
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:
\(10A=\frac{10^{2015}+20200}{10^{2015}+2020}=1+\frac{18180}{10^{2015}+2020}\)
\(10B=\frac{10^{2016}+20200}{10^{2016}+2020}=1+\frac{18180}{10^{2016}+2020}\)
Vì \(10^{2016}+2020>2^{2015}+2020\)
=> \(\frac{18180}{10^{2016}+2020}< \frac{18180}{10^{2015}+2020}\)
=> \(1+\frac{18180}{10^{2016}+2020}< 1+\frac{18180}{10^{2015}+2020}\)
=> 10B < 10A
=> B<A
\(A=\frac{10^{2014}+2020}{10^{2015}+2020}\)\(< \) \(B=\frac{10^{2015}+2020}{10^{2016}+2020}\)
chúc bạn học tốt
study well