Áp dụng tính chất \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+m}{b+m}\left(m\in N\right)\)

Ta có: \(\frac{10^{2019}-1}{10^{2020}-1}< \frac{10^{2019}-1+11}{10^{2020}-1+11}=\frac{10^{2019}+10}{10^{2020}+10}=\frac{10.\left(10^{2018}+1\right)}{10.\left(10^{2019}+1\right)}=\frac{10^{2018}+1}{10^{2019}+1}\)

\(\Rightarrow\frac{10^{2019}-1}{10^{2020}-1}< \frac{10^{2018}+1}{10^{2019}+1}\)

Đặt \(A=\frac{10^{2019}-1}{10^{2020}-1}\)

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

Dễ thấy \(A< 1\)

Áp dụng kết quả bài trên nếu \(\frac{a}{b}< 1\)thì \(\frac{a+m}{b+m}>\frac{a}{b}\)với m>0

Vậy \(A=\frac{10^{2019}-1}{10^{2020}-1}< \frac{\left[10^{2019}-1\right]+11}{\left[10^{2020}-1\right]+11}=\frac{10^{2019}+10}{10^{2020}+10}\)

\(A< \frac{10\left[10^{2018}+1\right]}{10\left[10^{2019}+1\right]}=\frac{10^{2018}+1}{10^{2019}+1}=B\)

Do đó : A<B

#)Giải :

Đặt \(A=\frac{10^{2019}-1}{10^{2020}-1}\)

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

\(\Rightarrow10A=\frac{10^{2020}-10}{10^{2020}-1}=\frac{10^{2020}-1+9}{10^{2020}-1}=\frac{10^{2020}-1}{10^{2020}-1}+\frac{9}{10^{2020}-1}=1+\frac{9}{10^{2020}-1}\)

\(\Rightarrow10B=\frac{10^{2019}+10}{10^{2019}+1}=\frac{10^{2019}+1+9}{10^{2019}+1}=\frac{10^{2019}+1}{10^{2019}+1}+\frac{9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)

Dễ thấy \(10^{2020}-1>10^{2019}+1\)

\(\Rightarrow\frac{9}{10^{2020}-1}>\frac{9}{10^{2019}+1}\Rightarrow1+\frac{9}{10^{2020}-1}>1+\frac{9}{10^{2019}+1}\Rightarrow10A>10B\Rightarrow A>B\)

\(\Rightarrow\frac{10^{2019}-1}{10^{2020}-1}>\frac{10^{2018}+1}{10^{2019}+1}\)

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 !

Ta có : \(A=\frac{10^{2018}+1}{10^{2019}+1}\)và \(B=\frac{10^{2020}+1}{10^{2020}+1}\)

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

Xét : \(B=\frac{10^{2020}+1}{10^{2020}+1}=1\)(2)

Từ (1) ; (2) Suy ra đpcm 

a) Ta có : \(\frac{-60}{12}=-5=-\frac{25}{5}\)

\(-0,8=-\frac{8}{10}=-\frac{4}{5}\)

Mà -25 < -4 nên \(\frac{-25}{5}< \frac{-4}{5}\)=> \(\frac{-60}{12}< -0,8\)

b) Ta có : \(\frac{2020}{2019}=1+\frac{1}{2019}\)

\(\frac{2021}{2020}=1+\frac{1}{2020}\)

Vì \(\frac{1}{2019}>\frac{1}{2020}\)nên \(\frac{2020}{2019}>\frac{2021}{2020}\)

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

\(\frac{10^{2019}+1}{10^{2020}+1}=\frac{10\left(10^{2019}+1\right)}{10^{2020}+1}=\frac{10^{2020}+10}{10^{2020}+1}=\frac{10^{2020}+1+9}{10^{2020}+1}=1+\frac{9}{10^{2020}+1}\)(2)

Đến đây tự so sánh rồi nhé

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

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

\(\Rightarrow10M=\frac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\frac{10^{2019}+1+9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)

\(N=\frac{10^{2019}+1}{10^{2020}+1}\)

\(\Rightarrow10N=\frac{10\left(10^{2019}+1\right)}{10^{2020}+1}=\frac{10^{2020}+1+9}{10^{2020}+1}=1+\frac{9}{10^{2020}+1}\)

Ta co: \(\frac{9}{10^{2019}+1}>\frac{9}{10^{2020}+1}\) ma \(1=1\)

\(\Rightarrow1+\frac{9}{10^{2019}+1}>1+\frac{9}{10^{2020}+1}\)

\(\Rightarrow10M>10N\)

\(\Rightarrow M>N\)

Ta có \(b-a=9.10^{2019}-\dfrac{9}{10^{2021}}>0\Rightarrow b>a\).