\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}=\frac{1+5\left(1 +5+5^2+...+5^8\right)}{1+5+5^2+...+5^8}=5+\frac{1}{1+5+5^2+...+5^8} \)

\(B=\frac{1+3+3^2+....+3^9}{1+3+3^2+....+3^8}=\frac{1+3\left(1+3+3^2+....+3^8\right)}{1+3+3^2+....+3^8}=3+\frac{1}{1+3+3^2+....+3^8}\)

\(=5+\frac{1}{1+3+3^2+....+3^8}-2\)  

Có: \(\frac{1}{1+5+5^2+...+5^8}>0\)              và      \(\frac{1}{1+3+3^2+....+3^8}-2< 0\)

\(\Rightarrow A>B\)

Mình chỉ hướng dẫn bạn thôi nhé!

1. Nhân M vs 10 và N vs 10

2.Tách 10M thành 1 + ... và N cũng vậy.

3.So sánh.

Vậy nhé!

CHÚ Ý: bài toán sau: với \(\frac{a}{b}< 1,\)\(\frac{a}{b}< \frac{a+m}{b+m}\)

\(\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+14}{19^{32}+5+14}=\frac{19^{31}+19}{19^{32}+19}< \frac{19\left(19^{30}+1\right)}{19\left(19^{31}+1\right)}< \frac{19^{30}+1+4}{19^{31}+1+4}=\frac{19^{30}+5}{19^{31}+5}\)

Rõ ràng N<1 nên theo N, nếu \(\frac{a}{b}< 1\Rightarrow\frac{a+n}{b+n}>\frac{a}{b}\)

=> \(\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5.19}{19^{31}+5.19}=\frac{19\left(19^{30}+5\right)}{19\left(19^{31}+5\right)}\)\(=\frac{19^{30}+5}{19^{31}+5}\)

Vậy N<M

mk nghĩ là A>B

a, \(B=\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+90}{19^{32}+5+90}=\frac{19^{31}+95}{19^{32}+95}=\frac{19\left(19^{30}+5\right)}{19\left(19^{31}+5\right)}=\frac{19^{30}+5}{19^{31}+5}=A\)

b, Ta có: \(\frac{1}{A}=\frac{2^{20}-3}{2^{18}-3}=\frac{2^2.\left(2^{18}-3\right)+9}{2^{18}-3}=4+\frac{9}{2^{18}-3}\)

\(\frac{1}{B}=\frac{2^{22}-3}{2^{20}-3}=\frac{2^2\left(2^{20}-3\right)+9}{2^{20}-3}=4+\frac{9}{2^{20}-3}\)

Vì \(\frac{9}{2^{18}-3}>\frac{9}{2^{20}-3}\)\(\Rightarrow\frac{1}{A}>\frac{1}{B}\Rightarrow A< B\)

c,  Câu hỏi của truong nguyen kim