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

\(\Rightarrow A< B\)

a) Ta có : B = \(\frac{9^{19}+1}{9^{20}+1}\)< \(\frac{9^{19}+1+8}{9^{20}+1+8}\)= \(\frac{9^{19}+9}{9^{20}+9}\)= \(\frac{9\left(9^{18}+1\right)}{9\left(9^{19}+1\right)}\)= \(\frac{9^{18}+1}{9^{19}+1}\)= A

                                                       Vậy A > B

b) Ta có : B = \(\frac{10^{2018}-1}{10^{2019}-1}\)> \(\frac{10^{2018}-1-9}{10^{2019}-1-9}\)= \(\frac{10^{2018}-10}{10^{2019}-10}\)= \(\frac{10\left(10^{2017}-1\right)}{10\left(10^{2018}-1\right)}\)= \(\frac{10^{2017}-1}{10^{2018}-1}\)= A

                                                                         Vậy A < B.

                    NHỚ K CHO MK VỚI NHÉ !!!!!!!!

a A lon hon B

a) 

\(9A=\frac{9^{19}+9}{9^{19}+1}=\frac{9^{19}+1+8}{9^{19}+1}=1+\frac{8}{9^{19}+1}\)

\(9A=\frac{9^{20}+9}{9^{20}+1}=\frac{9^{20}+1+8}{9^{20}+1}=1+\frac{8}{9^{20}+1}\)

ta thấy \(9^{19}+1< 9^{20}+1\Rightarrow\frac{8}{9^{19}+1}>\frac{8}{9^{20}+1}\)

\(\Rightarrow9A>9B\Rightarrow A>B\)

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

A>B do A>4 cònB<4

Ai biết thì giúp mình nha cần gấp

\(A=\frac{10^{2017}}{10^{2018+1}}=\frac{10^{2017}}{10^{2019}}=\frac{1}{10^2}\) 

Tương Tự với \(B=\frac{1}{10^2}\)

\(\Rightarrow A=B\)

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chuc ban hok tot

Ta có : \(\frac{10^{2019}-1}{10^{2018}-1}< \frac{10^{2019}-1+11}{10^{2018}-1+11}=\frac{10^{2019}+10}{10^{2018}+10}=\frac{10\left(10^{2018}+1\right)}{10\left(10^{2017}+1\right)}=\frac{10^{2018}+1}{10^{2017}+1}\)

Vậy     \(\frac{10^{2019}-1}{10^{2018}-1}< \frac{10^{2018}+1}{10^{2017}+1}\)

trả lời luôn câu hỏi thứ 2 của minhf nhé