\(M=\dfrac{10^{20}+1}{10^{19}+1}\)

\(N=\dfrac{10^{21}+1}{10^{20}+1}< \dfrac{10^{21}+1+9}{10^{20}+1+9}=\dfrac{10^{21}+10}{10^{20}+10}=\dfrac{10\left(10^{20}+1\right)}{10\left(10^{19}+1\right)}=\dfrac{10^{20}+1}{10^{19}+1}=M\)

\(\Rightarrow N< M\)

Sai rồi

M = \(\dfrac{10^{20}+1}{10^{19}+1}\) = 10 - \(\dfrac{9}{10^{19}+1}\) ;      N = \(\dfrac{10^{21}+1}{10^{20}+1}\) = 10 - \(\dfrac{9}{10^{20}+1}\)

Vì  \(\dfrac{9}{10^{19}+1}\) > \(\dfrac{9}{10^{20}+1}\) 

⇒ M < N (phân số nào có phần bù lớn hơn thì phân số đó nhỏ hơn)

\(M=\dfrac{10^{20}+1}{10^{19}+1}\)

\(N=\dfrac{10^{21}+1}{10^{20}+1}< \dfrac{10^{21}+1+9}{10^{20}+1+9}=\dfrac{10^{21}+10}{10^{20}+10}=\dfrac{10\left(10^{20}+1\right)}{10\left(10^{19}+1\right)}=\dfrac{10^{20}+1}{10^{19}+1}=M\)

\(\Rightarrow N< M\)

N < M.

Áp dụng \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+c}{b+c}\) (a;b;c \(\in\) N*)

Ta có:

\(B=\frac{10^{20}+1}{10^{21}+1}< \frac{10^{20}+1+9}{10^{21}+1+9}=\frac{10^{20}+10}{10^{21}+10}\)

\(B< \frac{10.\left(10^{19}+1\right)}{10.\left(10^{20}+1\right)}=\frac{10^{19}+1}{10^{20}+1}=A\)

=> A > B

Ta dùng bất đẳng thức\(\frac{a}{b}<\frac{a+n}{b+n}\left(n\ne0\right)\)

Ta có \(B=\frac{10^{20}+1}{10^{21}+1}<\frac{10^{20}+1+9}{10^{21}+1+9}<\frac{10^{20}+10}{10^{21}+10}<\frac{10\left(10^{19}+1\right)}{10\left(10^{20}+1\right)}\) 

               \(<\frac{10^{19}+1}{10^{20}+1}\)

Vậy \(A>B\)

bạn trần quang đài sai vài chỗ nhỏ đáng kể đấy

a=b moi dung chu

Ta có:\(B=\frac{10^{20}+1}{10^{21}+1}< 1\Rightarrow B=\frac{10^{20}+1}{10^{21}+1}< \frac{10^{20}+1+9}{10^{21}+1+9}=\frac{10^{20}+10}{10^{21}+10}=\frac{10\left(10^{19}+1\right)}{10\left(10^{20}+1\right)}=\frac{10^{19}+1}{10^{20}+1}=A\)

=> A > B

đúng rồi mk cũng làm như vậy

\(B=\frac{10^{20}+1}{10^{21}+1}< 1\)

NÊN \(\frac{10^{20}+1}{10^{21}+1}< \frac{10^{20}+1+9}{10^{21}+1+9}=\frac{10^{20}+10}{10^{21}+10}=\frac{10.\left(10^{19}+1\right)}{10.\left(10^{20}+1\right)}=\frac{10^{19}+1}{10^{20}+1}=A\)

VẬY B<A

Ta có:

\(B=\frac{20^{10}-1}{20^{10}-3}>1\)

=> Ta có: \(B=\frac{20^{10}-1}{20^{10}-3}>\frac{20^{10}-1-2}{20^{10}-3-2}=\frac{20^{11}+1}{20^{10}+1}=A\)

\(\Rightarrow A< B\)

\(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\) 

\(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)

vì \(\frac{2}{20^{10}-1}\) < \(\frac{2}{20^{10}-3}\) 

=> 1 + \(\frac{2}{20^{10}-1}\) < 1+\(\frac{2}{20^{10}-3}\)

hay A< B