so sánh
2011.2013+2012.2014
\(2012^2+2013^2-2\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
Y
23 tháng 7 2018
a) \(2011.2013+2012.2014\)
\(=\left(2012-1\right)\left(2012+1\right)+\left(2013-1\right)\left(2013+1\right)\)
\(=2012^2-1+2013^2-1\)
\(=2012^2+2013^2-2\)
\(\Rightarrow2011.2013+2012.2014=2012^2+2013^2-2\)
b) \(\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9+1\right)\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^2-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^4-1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^8-1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^{16}-1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^{32}-1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^{64}-1\right)\)
\(=\dfrac{9^{64}-1}{10}\)
Ta có: \(9^{64}-1=\dfrac{10\left(9^{64}-1\right)}{10}\)
Mà \(\dfrac{10\left(9^{64}-1\right)}{10}>\dfrac{9^{64}-1}{10}\)
\(\Rightarrow\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)< 9^{64}-1\)
c) Ta có:
\(\dfrac{x^2-y^2}{x^2+xy+y^2}=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2-xy}\left(1\right)\)
Vì x>y>0, ta có:
\(\dfrac{x-y}{x+y}=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}\left(2\right)\)
Vì x>y>0 nên \(\left(x+y\right)^2-xy< \left(x+y\right)^2\left(3\right)\)
Từ (1), (2) và (3) suy ra:
\(\dfrac{x-y}{x+y}< \dfrac{x^2-y^2}{x^2+xy+y^2}\)
PT
18 tháng 9 2018
a) Ta có:
\(2011.2013+2012.2014\)
\(=\left(2012-1\right)\left(2012+1\right)+\left(2013-1\right)\left(2013+1\right)\)
\(=2012^2-1+2013^2-1\)
\(=2012^2+2013^2-2\)
Vậy 2011.2013+2012.2014 = 20122 + 20132 - 2
TN
2
BS
30 tháng 6 2021
A= 2013. ( 2021 + 1 ) = 2013 . 2012 + 2013
B = 2012 . 2014 = 2012 . ( 2013 + 1 ) = 2012 . 2013 + 2012
Vì 2013 > 2012 ==> A > B
NT
0
2011.2013+2012.2014
=(2013-2).2013+2012.(2012+2)
=20132-4026+20122+4024
=20132+20122+(-4026+4024)
=20132+20122-2
Ta có:\(2011.2013+2012.2014\)
\(=\left(2013-2\right).2013+\left(2012+2\right).2012\)
\(=2013^2-4026+2012^2+4024\)
\(=2012^2+2013^2-2\)
nên hai phép tính trên bằng nhau.