Nhân cả hai tử của \(A\)và \(B\)với 2 , ta được :

\(10A=10.\left(\frac{10^{2016}+1}{10^{2017}+1}\right)=\frac{10^{2017}+1+9}{10^{2017}+1}=1+\frac{9}{2^{2017}+1}\)

\(10B=10\left(\frac{10^{2017}+1}{10^{2018}+1}\right)=\frac{10^{2018}+10}{10^{2018}+1}=\frac{10^{2018}+1+9}{10^{2018}}=1+\frac{9}{10^{2018}+1}\)

Vì \(1=1;9=9\)

\(\Rightarrow\)Ta so sánh mẫu , ta có:

\(10^{2017}< 10^{2018}\)

\(\Rightarrow10^{2017}+1< 10^{2018}+1\)

\(\Rightarrow1+\frac{9}{10^{2017}+1}>1+\frac{9}{10^{2018}+1}\)

\(\Rightarrow10A>10B\)

Hay \(A>B\)

Ta có :  \(A=\frac{10^{2016}+1}{10^{2017}+1}\) 

Suy ra  \(10A=\frac{10^{2017}+10}{10^{2017}+1}\) 

Suy ra  \(10A=1+\frac{9}{10^{2017}+1}\) 

Ta lại có : \(B=\frac{10^{2017}+1}{10^{2018}+1}\) 

Suy ra : \(10B=\frac{10^{2018}+10}{10^{2018}+1}\) 

Suy ra : \(10B=1+\frac{9}{10^{2018}+1}\) 

Vì  \(\frac{9}{10^{2017}+1}>\frac{9}{10^{2018}+1}\) 

Nên  \(1+\frac{9}{10^{2017}+1}>1+\frac{9}{10^{2018}+1}\) 

Suy ra \(10A>10B\) 

Suy ra \(A>B\)

\(B< \frac{10^{2017}+1+9}{10^{2018}+1+9}=\frac{10^{2017}+10}{10^{2018}+10}=\frac{10\left(10^{2016}+1\right)}{10\left(10^{2017}+1\right)}=\frac{10^{2016}+1}{10^{2017}+1}=A\)

vậy A > B

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Tường đây

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 : 

\(B=\frac{10^{2018}+1}{10^{2017}+1}>\frac{10^{2018}+1+9}{10^{2018}+1+9}=\frac{10^{2018}+10}{10^{2018}+10}=\frac{10\left(10^{2017}+1\right)}{10\left(10^{2016}+1\right)}=\frac{10^{2017}+1}{10^{2016}+1}=A\)

\(\Rightarrow\)\(B>A\) hay \(A< B\)

Vậy \(A< B\)

Chúc bạn học tốt ~ 

Ta có: \(\hept{\begin{cases}A=\frac{10^{2016}+1}{10^{2017}+1}\\B=\frac{10^{2017}+1}{10^{2018}+1}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}10A=\frac{10^{2017}+10}{10^{2017}+1}=\frac{10^{2017}+1+9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\\10B=\frac{10^{2018}+10}{10^{2018}+1}=\frac{10^{2018}+1+9}{10^{2018}+1}=1+\frac{9}{10^{2018}+1}\end{cases}}\)

Vì \(\frac{9}{10^{1017}+1}>\frac{9}{10^{2018}+1}\)

nên \(10A>10B\Rightarrow A>B\)

\(A=\frac{10^{2016}+1}{10^{2017}+1}\Rightarrow10A=\frac{10\cdot(10^{2016}+1)}{10^{2017}+1}=\frac{10^{2017}+10}{10^{2017}+1}\)

\(A=\frac{10^{2017}+1+9}{10^{2017}+1}=\frac{10^{2017}+1}{10^{2017}+1}+\frac{9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\)

Vì \(10^{2016}+1< 10^{2017}+1\)

\(\Rightarrow\frac{9}{10^{2016}+1}>\frac{9}{10^{2017}+1}\)

\(\Rightarrow\)\(1+\frac{9}{10^{2016}+1}>1+\frac{9}{10^{2017}+1}\)

....

A= \(\frac{10^{2016}+1}{10^{2016}+1}=\frac{10^{2016}+1}{10\cdot10^{2016}+1}=\frac{1}{10}\cdot\frac{10^{2016}+1}{10^{2016}+1}=\frac{1}{10}\)(1)

B=\(\frac{10^{2017}+1}{10^{2018}+1}=\frac{10^{2017}+1}{10\cdot10^{2017}+1}=\frac{1}{10}\cdot\frac{10^{2017}+1}{10^{2017}+1}=\frac{1}{10}\)(2)

Từ (1) và (2) \(\Rightarrow\)A=B

Ta có :

A = \(\frac{10^{2017}+1}{10^{2018}+1}\)< 1 => A < \(\frac{10^{2017}+1+9}{10^{2018}+1+9}\)= \(\frac{10^{2017}+10}{10^{2018}+10}\)= \(\frac{10^{2016}+1}{10^{2017}+1}\)= B

Vậy A < B

A<B. lời giải thích khó viết lắm nên bạn tự tìm cách làm nhé

Ta có: \(B=\frac{10^2\left(10^{2017}+1\right)}{10^2\left(10^{2016}+1\right)}=\frac{10^{2019}+1+99}{10^{2018}+1+99}\)

Do phân số \(A=\frac{10^{2019}+1}{10^{2018}+1}>1\).Áp dụng BĐT \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+m}{b+m}\left(m>0\right)\).

Ta có: \(A=\frac{10^{2019}+1}{10^{2018}+1}>\frac{10^{2019}+1+99}{10^{2018}+1+99}=B\)

Vậy \(A>B\)

C/m BĐT phụ nè: \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+m}{b+m}\left(m>0\right)\)

\(\Leftrightarrow a\left(b+m\right)>b\left(a+m\right)\)

\(\Leftrightarrow ab+am>ab+bm\)

\(\Leftrightarrow am>bm\Leftrightarrow a>b\) (đúng,do \(\frac{a}{b}>1\))

\(+)A=\frac{10^{2016}+2018}{10^{2017}+2018}\)

\(10A=\frac{10^{2017}+20180}{10^{2017}+2018}=1+\frac{18162}{10^{2017}+2018}\left(1\right)\)

\(+)10B=\frac{10^{2018}+20180}{10^{2018}+2018}=1+\frac{18162}{10^{2018}+2018}\left(2\right)\)

Từ (1),(2)=> \(\frac{18162}{10^{2017}+2018} >\frac{18162}{10^{2018}+2018}\)

=> 10A>10B

=>A>B

k đúng cho mình đi, mình giải cho.