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6 tháng 4 2019

2011.2013+2012.2014

=(2013-2).2013+2012.(2012+2)

=20132-4026+20122+4024

=20132+20122+(-4026+4024)

=20132+20122-2

6 tháng 4 2019

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.

23 tháng 7 2018

\(2011.2013+2012.2014\)

\(=\left(2012-1\right)\left(2012+1\right)+\left(2013-1\right)\left(2013+1\right)\)

\(=2012^2-1+2013^2-1\)

\(=2012^2+2013^2-2\)

\(\Rightarrow2011.2013+2012.2014=2012^2+2013^2-2\)

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: \(9^{64}-1=\dfrac{10\left(9^{64}-1\right)}{10}\)

\(\dfrac{10\left(9^{64}-1\right)}{10}>\dfrac{9^{64}-1}{10}\)

\(\Rightarrow\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)< 9^{64}-1\)

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:

\(\dfrac{x-y}{x+y}< \dfrac{x^2-y^2}{x^2+xy+y^2}\)

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

30 tháng 4 2018

https://hoc24.vn/hoi-dap/question/598367.html

30 tháng 9 2015

Siêu Trí Tuệ sai

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

9 tháng 9 2018

\(=2013^4-\left(2012\cdot2014\right)\left(2013^2+1\right)\\ =2013^4-\left(2013^2-1\right)\left(2013^2+1\right)\\ =2013^4-\left(2013^4-1\right)\\ =1\)