a) Ta có A = \(\frac{2^{2018}+1}{2^{2019}+1}\)

=> 2A = \(\frac{2^{2019}+2}{2^{2019}+1}=1+\frac{1}{2^{2019}+1}\)

Lại có B = \(\frac{2^{2017}+1}{2^{2018}+1}\)

=> 2B = \(\frac{2^{2018}+2}{2^{2018}+1}=\frac{2^{2018}+1+1}{2^{2018}+1}=1+\frac{1}{2^{2018}+1}\)

Vì \(\frac{1}{2^{2018}+1}>\frac{1}{2^{2019}+1}\Rightarrow1+\frac{1}{2^{2018}+1}>1+\frac{1}{2^{2019}+1}\Rightarrow2B>2A\Rightarrow B>A\)

ta có :

\(A=\frac{10^{2019}+1}{10^{2018}+1}=\frac{10^{2018}.10+1}{10^{2018}+1}=\frac{10}{10^{2018}+1}\)

\(B=\frac{10^{2018}+1}{10^{2017}+1}=\frac{10^{2017}.10+1}{10^{2017}+1}=\frac{10}{10^{2017}+1}\)

Do \(10^{2017}+1< 10^{2018}+1\Rightarrow\frac{10}{10^{2017}+1}>\frac{10}{10^{2018}+1}\)

\(\Rightarrow A< B\)

Ta thấy \(B=\frac{10^{2020}+1}{10^{2020}+1}=1\)

            \(A=\frac{10^{2018}+1}{10^{2019}+1}< 1\)

\(\Rightarrow A< B\)

Vậy \(A< B.\)

Bạn có chắc là đề đúng không?

                             Bài giải

A < 1 ; B = 1 => A < B

Nếu đề bạn sai thì vào câu hỏi tương tự là có !

Gà

Ta có:

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

10B=\(\frac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\frac{10^{2019}+10}{10^{2019}+1}=\frac{10^{2019}+1}{10^{2019}+1}+\frac{9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)

do 1=1 và \(\frac{9}{10^{2018}+1}>\frac{9}{10^{2019}+1}\)

\(\Rightarrow\)A>B

Vậy A>B

chúc bạn học tốt!

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 \(\frac{a}{b}>1\))

\(M=\frac{2018^{2018}+1}{2019^{2019}+1}\)

\(\Leftrightarrow2M=1+\frac{2017}{2018^{2019}+1}\)

\(N=\frac{2018^{2019}-2}{2018^{2020}-2}\)

\(\Leftrightarrow2N=1-\frac{4034}{2018^{2020}-2}\)

Nhận thấy :  \(1+\frac{2017}{2018^{2019}+1}>1-\frac{4034}{2018^{2020}-2}\Leftrightarrow2M>2N\Leftrightarrow M>N\)

Từ đề bài, ta suy ra:

So sánh hai biểu thức

\(M=\left(2018^{2018}+1\right)\cdot\left(2018^{2020}-2\right)\)(1)

\(N=\left(2018^{2019}-2\right)\cdot\left(2018^{2019}+1\right)\)(2)

Xét biểu thức M và N, ta suy ra:

\(M=\left(2018^{2019}-2017\right)\cdot\left(2019^{2019}+2016\right)\)

\(N=\left(2018^{2019}-2017\right)\cdot\left(2018^{2018}-2016\right)\)

Nhận thấy (2019²⁰¹⁹+2016)>(2018²⁰¹⁸-2016) nên M>N

Vậy M>N.

P/s:Mình đây không phải top 10 tuần nên bài có thể sai sót, mong bạn tham khảo:)))