Ta có:

\(a^{2010}+b^{2010}+a^{2012}+b^{2012}\)

\(=\left(a^{2010}+a^{2012}\right)+\left(b^{2010}+b^{2012}\right)\ge2a^{2011}+2b^{2011}\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}a^{2010}=a^{2012}\\b^{2010}=b^{2012}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=1\\b=1\end{cases}}\)

\(\Rightarrow a^{2013}+b^{2013}=2\)

Vậy \(S=2\)

thank ban nha

Ta có:

\(\frac{2011}{2012}=1-\frac{1}{2012}\)

\(\frac{2012}{2013}=1-\frac{1}{2013}\)

\(\frac{2013}{2014}=1-\frac{1}{2014}\)

Do \(\frac{1}{2012}>\frac{1}{2013}>\frac{1}{2014}\)=> \(-\frac{1}{2012}< -\frac{1}{2013}< -\frac{1}{2014}\)

=> \(1-\frac{1}{2012}< 1-\frac{1}{2013}< 1-\frac{1}{2014}\)

=> \(\frac{2011}{2012}< \frac{2012}{2013}< \frac{2013}{2014}\)

1) 1

2)Ta có: 2011 x 2013 + 2012 x 2014 =8100311

2012² + 2013² - 2 =8100311 . 

Vậy ta đã thấy 2 số bằng nhau

Kết luận : 2011 x 2013 + 2012 x 2014 = 2012²+ 2013² - 2

1, \(B=3^{24}-\left(27^4+1\right)\left(9^6-1\right)\)

\(=\left(3^{12}\right)^2-\left(3^{12}+1\right)\left(3^{13}-1\right)\)

\(=\left(3^{12}\right)^2-\left[\left(3^{12}\right)^2-1\right]\)

\(=\left(3^{12}\right)^2-\left(3^{12}\right)^2+1\)

\(=1\)

Vậy \(B=1\)

A=B đúng rồi đó nha bạn

nhờ giải thích dùm mình làm bt mà bạn

1) 1

2) hai số bằng nhau

Ta có: A= 2011.2013 + 2013.2015 

            = (2012 - 1)(2012 + 1) + (2014 - 1)(2014 + 1)

            = 2012^2 + 2012 - 2012 - 1 + 2014^2 +2014 - 2014 - 1

            = 2012^2 + 2014^2 - 2

            =            B            - 2  

     Vì B - 2 < B nên A < B

A = 2014 . (2015+1) = 2014 . 2015 + 2014

B= 2015^2 = 2015(2014 + 1) = 2014 . 2015 +2015

Vì 2014<2015 => 2014.2015 + 2014 < 2014.2015  +2015 

=> A< B

Vậy A<B

A = 2014 . (2015+1) = 2014 . 2015 + 2014

B= 2015^2 = 2015(2014 + 1) = 2014 . 2015 +2015

Vì 2014<2015 => 2014.2015 + 2014 < 2014.2015  +2015 

=> A< B

Vậy A<B

\(201^2=\left(200+1\right)^2=200^2+2.200.1+1^2=40000+400+1=40401\)

\(498^2=\left(500-2\right)^2=500^2-2.500.2+2^2=250000-2000+4=248004\)

\(93.107=\left(100-7\right)\left(100+7\right)=100^2-7^2=10000-49=9951\)

\(2016^2-2015.2017=2016^2-\left(2016-1\right)\left(2016+1\right)=2016^2-2016^2+1^2=1\)