A- 1 = \(\frac{10^{2015}-1-\left(10^{2016}-1\right)}{10^{2016}-1}=\frac{-9.10^{2015}}{10^{2016}-1}=\frac{-90.10^{2014}}{10^{2016}-1};\)

B- 1 = \(\frac{10^{2014}+1-\left(10^{2015}+1\right)}{10^{2015}+1}=\frac{-9.10^{2014}}{10^{2015}+1};\)

xét \(\frac{A-1}{B-1}=\frac{-90.10^{2014}}{10^{2016}-1}:\frac{-9.10^{2014}}{10^{2015}+1}=\frac{10\left(10^{2015}+1\right)}{10^{2016}-1}=\frac{10^{2016}+10}{10^{2016}-1}>1\)

=> A-1 > B-1 => A > B

A=10^2014+1/10^2015+1

10A=10^2015+10/10^2015+1

10A=10^2015+1+9/10^2015+1

10A=1+(9/10^2015+1)(1)

B làm tương tự (2)

Từ (1); (2)

Suy ra 10A>10B

Suy ra A>B

Vậy........

Vi B < 1 nen ta co : 

 \(B=\frac{10^{2015}+1}{10^{2016}+1}< \frac{10^{2015}+1+9}{10^{2016}+1+9}\)

\(\Rightarrow B< \frac{10^{2015}+10}{10^{2016}+10}=\frac{10\left(10^{2014}+1\right)}{10\left(10^{2015}+1\right)}=A\)

Vay \(B< A\)

\(A=\frac{10^{2015}-1}{10^{2016}^{ }-1}=\frac{10^{2015}}{10^{2016}}=\frac{1}{1},B=\frac{10^{2014}-1}{10^{2015}-1}=\frac{10^{2014}}{10^{2015}}=\frac{1}{1}A=B\Rightarrow\)

A=\(\frac{10^{2015}+1}{10^{2016}+1}\)=>10A=\(\frac{10.\left(10^{2015}+1\right)}{10^{2016}+1}\)= \(\frac{10^{2016}+10}{10^{2016}+1}\)=\(\frac{\left(10^{2016}+1\right)+9}{10^{2016}+1}\)=\(\frac{10^{2016}+1}{10^{2016}+1}+\frac{9}{10^{2016}+1}\)=1+\(\frac{9}{10^{2016}+1}\)

B=\(\frac{10^{2016}+1}{10^{2017}+1}\)=>10B=\(\frac{10.\left(10^{2016}+1\right)}{10^{2017+1}}=\frac{10^{2017}+10}{10^{2017}+1}\)= \(\frac{\left(10^{2017}+1\right)+9}{10^{2017}+1}\)=\(\frac{10^{2017}+1}{10^{2017}+1}+\frac{9}{10^{2017}+1}\)= 1+\(\frac{9}{10^{2017}+1}\)

Vì \(10^{2016}+1< 10^{17}+1\)=>\(\frac{9}{10^{2016}+1}\)>\(\frac{9}{10^{2017}+1}\)nên \(1+\frac{9}{10^{2016}+1}>1+\frac{9}{10^{2017}+1}\)=>10A>10B

Vậy A>B

Cảm ơn bạn nhìu nhé.

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 \(A< \frac{1}{4}\)

Ta có công thức : 

\(\frac{a}{b}>\frac{a+c}{b+c}\)\(\left(\frac{a}{b}>1;a,b,c\inℕ^∗\right)\)

Áp dụng vào ta có : 

\(B=\frac{10^{2016}+1}{10^{2015}+1}>\frac{10^{2016}+1+9}{10^{2015}+1+9}=\frac{10^{2016}+10}{10^{2015}+10}=\frac{10\left(10^{2015}+1\right)}{10\left(10^{2014}+1\right)}=\frac{10^{2015}+1}{10^{2014}+1}=A\)

\(\Rightarrow\)\(B>A\) hay \(A< B\)

Vậy \(A< B\)

Chúc bạn học tốt ~ 

Xét  \(A=\frac{10^{2014}+2016}{10^{2015}+2016}\Rightarrow10A=\frac{10^{2015}+20160}{10^{2015}+2016}=\frac{10^{2015}+2016+18144}{10^{2015}+2016}=1+\frac{18144}{10^{2015}+2016}\)

Xét \(B=\frac{ 10^{2015}+2016}{10^{2016}+2016}\Rightarrow10B=\frac{10^{2016}+20160}{10^{2016}+2016}=\frac{10^{2016}+2016+18144}{10^{2016}+2016}=1+\frac{18144}{10^{2016}+2016}\)

Có \(\frac{18144}{10^{2015}+2016}>\frac{18144}{10^{2016}+2016}\)

\(\Rightarrow10A>10B\Leftrightarrow A>B\)

cảm ơn bạn nha