So sánh: A=\(\frac{10^{2014}+2020}{10^{2015}+2020}\)và B=\(\frac{10^{2015}+2020}{10^{2016}+2020}\)
Mình đang cần gấp các bạn giúp mình nhé ^_^
ko tính giá trị của biểu thức hãy so sánh A và B
A = 2010 . 2020 + 10 và B= 2015 . 2015 + 10
A= 2015 . 2020 _ 1 và B = 2010 .2025 _ 1
A = 2010 . 2020 + 10 và B = 2015 . 2015 + 10
A = 2010 . 2020 + 10
A = 2010 . ( 2015 + 5 ) + 10
A = 2010 . 2015 + 2010 . 5 + 10
B = 2015 . 2015 + 10
B = (2010 + 5) . 2015+ 10
B = 2010.2015 + 2015.5 + 10
Vì 2010.5 < 2015.5 nên A < B
A = 2015 . 2020 - 1
A = ( 2010 + 5 ) . 2020 - 1
A = 2010 . 2020 + 2020 . 5 - 1
B = 2010 . 2025 - 1
B = 2010 . ( 2020 + 5 ) - 1
B = 2010 . 2020 + 2010 . 5 - 1.
Vì 2020.5 > 2010.5 nên A > B.
( Dấu chấm là dấu nhân nha bạn )
so sánh
20202014+1/20202015+1 và 20202015-2/20202016-2
so sánh
a)
A=\(\frac{10^{2020}+1}{10^{2021}+1};B=\frac{10^{2021}+1}{10^{2022}+1}\)
b)
\(A=\frac{2019}{2020}+\frac{2020}{2021}\)và \(B=\frac{2019+2020}{2020+2021}\)
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
a) A=10^2020+1/10^2021+1 < 10^2020+1+9/10^2022+1+9 =
10.(10^2021+1)/10.(10^2022+1) = 10^2021+1/10^2022+1 = B
Vậy A < B.
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
cho mình cả s2 bạn ơi,thank bạn
So sánh A=\(\frac{2020^{10}+2}{2020^{11}+2}\)và B=\(\frac{2020^{11}+2}{2020^{12}+2}\)
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
So sánh A=\(\frac{2020^{10}+2}{2020^{11}+2}\)và B=\(\frac{2020^{11}+2}{2020^{12}+2}\)
So sánh A = \(\dfrac{10^{2014}+2016}{10^{2015}+2016}\) và B = \(\dfrac{10^{2015}+2016}{10^{2016}+2016}\) giúp mình nhanh với
\(10A=\dfrac{10^{2015}+2016+9\cdot2016}{10^{2015}+2016}=1+\dfrac{18144}{10^{2015}+2016}\)
\(10B=\dfrac{10^{2016}+9+18144}{10^{2016}+2016}=1+\dfrac{18144}{10^{2016}+2016}\)
mà \(\dfrac{18144}{10^{2015}+2016}>\dfrac{18144}{10^{2016}+2016}\)
nên A>B
\(A=-\frac{1}{2}\left(17,5-7,5\right)-\frac{2015}{2016}\left(2018-2\right)\)
=> \(A=-\frac{1}{2}\left(10\right)-\frac{2015}{2016}\left(2016\right)=-5-2015=-2020\)
Trả lời :
- 2 bn kia ở trong câu hỏi này có ai làm đúng đâu.
- Chúc bạn học tốt !
- Tk cho mk nha !
So sánh
a)\(\frac{-60}{12}\)và -0,8
b) \(\frac{2020}{2019}\)và \(\frac{2021}{2020}\)
c) \(\frac{10^{2018}+1}{10^{2019}+1}\)và \(\frac{10^{2019}+1}{10^{2020+1}}\)
a) Ta có : \(\frac{-60}{12}=-5=-\frac{25}{5}\)
\(-0,8=-\frac{8}{10}=-\frac{4}{5}\)
Mà -25 < -4 nên \(\frac{-25}{5}< \frac{-4}{5}\)=> \(\frac{-60}{12}< -0,8\)
b) Ta có : \(\frac{2020}{2019}=1+\frac{1}{2019}\)
\(\frac{2021}{2020}=1+\frac{1}{2020}\)
Vì \(\frac{1}{2019}>\frac{1}{2020}\)nên \(\frac{2020}{2019}>\frac{2021}{2020}\)
c) \(\frac{10^{2018}+1}{10^{2019}+1}=\frac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\frac{10^{2019}+10}{10^{2019}+1}=\frac{10^{2019}+1+9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)(1)
\(\frac{10^{2019}+1}{10^{2020}+1}=\frac{10\left(10^{2019}+1\right)}{10^{2020}+1}=\frac{10^{2020}+10}{10^{2020}+1}=\frac{10^{2020}+1+9}{10^{2020}+1}=1+\frac{9}{10^{2020}+1}\)(2)
Đến đây tự so sánh rồi nhé