\(\text{ ta có: }\frac{a}{b}=\frac{a.\left(b+2015\right)}{b.\left(b+2015\right)}=\frac{a.b+2015.a}{b^2+2015.b}\)

\(\frac{a+2015}{b+2015}=\frac{b.\left(a+2015\right)}{b.\left(b+2015\right)}=\frac{a.b+2015.b}{b^2+2015.b}\)

Nếu a>b thì :

\(a.b+2015.a>a.b+2015.b\Rightarrow\frac{a.b+2015.a}{b^2+2015.b}>\frac{a.b+2015.b}{b^2+2015.b}\)

hay \(\frac{a}{b}>\frac{a+2015}{b+2015}\)

Nếu a=b thì:

\(a.b+2015.a=a.b+2015.b\Rightarrow\frac{a.b+2015.a}{b^2+2015.b}=\frac{a.b+2015.b}{b^2+2015.b}\)

hay \(\frac{a}{b}=\frac{a+2015}{b+2015}\)

Nếu a<b thì:

a.b+2015.a<a.b+2015.b \(\Rightarrow\frac{a.b+2015.a}{b^2+2015.b}

\(\frac{2015}{2016}+\frac{2016}{2017}>\frac{\left(2015+2016\right)}{\left(2016+2017\right)}=\frac{2015}{2016+2017}+\frac{2016}{2016+2017}\)

ko bit

\(A=\frac{2015}{2016}+\frac{2016}{2017}=1-\frac{1}{2016}+1-\frac{1}{2017}>1\)

\(B=\frac{2015+2016}{2016+2017}< \frac{2016+2017}{2016+2017}=1\)

Suy ra \(A>B\).

có cách nhanh hơn

\(A=\frac{2015}{2016}+\frac{2016}{2017}\)   

\(=1-\frac{1}{2015}+1-\frac{1}{2016}\)    

\(=2-\left(\frac{1}{2015}+\frac{1}{2016}\right)>1\)   

\(B=\frac{2015+2016}{2016+2017}\)   

\(=\frac{4031}{4033}\)   

\(=1-\frac{2}{4033}< 1\)

Vậy A > B 

\(A=1+2+2^2+...+2^{2015}>2^{2015}=B\)

\(\Rightarrow A>B\)

P.s: đề sai đúng ko bạn :v

đề ko sai 

do bat sai

\(A=2^{2014.2015}.5^{2014.2015}\)

\(B=2^{2015.2014}.5^{2015.2014}\)

Vậy A = B

Haha , Việt làm sai đâu phải nhân đâu              

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 )