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.

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:

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

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

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

Siêu Trí Tuệ sai

A= 2013. ( 2021 + 1 ) = 2013 . 2012 + 2013

B = 2012 . 2014 = 2012 . ( 2013 + 1 ) = 2012 . 2013 + 2012

Vì 2013 > 2012 ==> A > B

a) 2011 . 2013 = 2011 . ( 2012 + 1 ) = 2011 . 2012 + 2011

2012² = 2012 . 2012 = ( 2011 + 1 ) . 2012 = 2011 . 2012 + 2012

Vì 2011 . 2012 + 2011 < 2011 . 2012 + 2012 nên 2011 . 2013 < 2012²

(3+4)²=3²+4²

Vì 3²+4²=3²+4² nên (3+4)²=3²+4²

2³⁰⁰=(2³)¹⁰⁰=8¹⁰⁰

3²⁰⁰=(3²)¹⁰⁰=9¹⁰⁰

Vì 8¹⁰⁰<9¹⁰⁰ nên 2³⁰⁰<3²⁰⁰

a, M=2011.2013=2011.(2012+1)=2011.2012+2011

N=2012^2=2012.(2011+1)=2012.2011+2012

=>M<N

b, M=2015^2015+2015^2016=2015^2015.(1+2015)=2015^2015.2016

N=2016^2016=2016^2015.2016

=>M<N

k cho k nha

A = 2011.2013

A = 2011.(2012+1)

A = 2011.2012+2011

B = 2012²

B = 2012.2012

B = (2011+1).2012

B = 2011.2012+2012

Vì 2011 < 2012

=> 2011.2012 + 2011 < 2011.2012 + 2012

=> A < B

Vậy a<b nha bạn