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Ta có: 

A=9²⁰¹³+1/9²⁰¹⁴+1

9A=9²⁰¹⁴+9/9²⁰¹⁴+1

    =(9²⁰¹⁴+1/9²⁰¹⁴+1)+(8/9²⁰¹⁴+1)

    =1+8/9²⁰¹⁴+1

B=9²⁰¹⁴+1/9²⁰¹⁵+1

9B=9²⁰¹⁵+9/9²⁰¹⁵+1

    =(9²⁰¹⁵+1/9²⁰¹⁵+1)+(8/9²⁰¹⁵+1)

    =1+8/9²⁰¹⁵+1

Vì 8/9²⁰¹⁴+1 > 8/9²⁰¹⁵+1 nên A>B

**** bạn

\(9A=\frac{9\left(9^{2014}+1\right)}{9^{2015+1}}=\frac{9^{2015}+9}{9^{2015}+1}=\frac{9^{2015}+1+8}{9^{2015}+1}=1+\frac{8}{9^{2015}+1}\)

\(9B=\frac{9\left(9^{2015}+1\right)}{9^{2016+1}}=\frac{9^{2016}+9}{9^{2016}+1}=\frac{9^{2016}+1+8}{9^{2016}+1}=1+\frac{8}{9^{2016}+1}\)

Ta thấy  \(9^{2016}+1>9^{2015}+1\Rightarrow\frac{8}{9^{2016}+1}<\frac{8}{9^{2015}+1}\)

suy ra 9A >9B

Vậy A > B

nghĩ đi nhé , giải ra thì k còn thú vị nữa , ^_^ còn k thì 15 ' sau pm mình giải cho

a,\(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)

\(=>5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)

\(=>5A-A=1-\frac{1}{5^{100}}=>A=\frac{1-\frac{1}{5^{100}}}{4}\)

b, Ta có \(1-\frac{1}{5^{100}}< 1=>\frac{1-\frac{1}{5^{100}}}{4}< \frac{1}{4}\)hay \(A< \frac{1}{4}\)

So sánh A và B biết :

A = a^a^9 = a^9.a

B = 9^9^a = 9^9.a

Đây là hai lũy thừa cùng số mũ 

mà a = 123456789 > 9 suy ra A > B  nhé

A > B nhá ^_^"

\(A=\frac{2014^{2015}+2}{2014^{2016}+9}\)

\(2014A=\frac{2014\left(2014^{2015}+2\right)}{2014^{2016}+9}=\frac{2014^{2016}+4028}{2014^{2016}+9}=\frac{\left(2014^{2016}+9\right)+4019}{2014^{2016}+9}=\frac{2014^{2016}+9}{2014^{2016}+9}+\frac{4019}{2014^{2016}+9}=1+\frac{4019}{2014^{2016}+9}\)

\(B=\frac{2014^{2016}+2}{2014^{2017}+9}\)

\(2014B=\frac{2014\left(2014^{2016}+2\right)}{2014^{2017}+9}=\frac{2014^{2017}+4028}{2014^{2017}+9}=\frac{2014^{2017}+9+4019}{2014^{2017}+9}=\frac{2014^{2017}+9}{2014^{2017}+9}+\frac{4019}{2014^{2017}+9}=1+\frac{4019}{2014^{2017}+9}\)

Ta thấy:

\(2014^{2016}+9< 2014^{2017}+9\)

\(\Rightarrow\frac{4019}{2014^{2016}+9}>\frac{4019}{2014^{2017}+9}\)

\(\Rightarrow1+\frac{4019}{2014^{2016}+9}>1+\frac{4019}{2014^{2017}+9}\)

\(\Rightarrow A>B\)

Vậy ....