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

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

\(=5+\frac{1}{1+3+3^2+....+3^8}-2\)  

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

\(\Rightarrow A>B\)

\(A=\frac{3^{10}+1}{3^9+1}=\frac{3^{10}+3-2}{3^9+1}=\frac{3\left(3^9+1\right)-2}{3^9+1}=3-\frac{2}{3^9+1}\)

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

Có \(3^9+1>3^8+1\)

\(\Rightarrow\frac{2}{3^9+1}< \frac{2}{3^8+1}\)

\(\Rightarrow3-\frac{2}{3^9+1}>3-\frac{2}{3^8+1}\)

\(\Rightarrow A>B\)

a, 28/63 và 27/63

b, -10/15 và -12/15

c, -1/5 và 5/5

mk nghĩ là A>B

a) Ta có:

+) \(\dfrac{1}{2}=\dfrac{3}{6}\)

+) \(\dfrac{1}{3}=\dfrac{2}{6}\)

+) \(\dfrac{2}{3}=\dfrac{4}{6}\)

=> \(\dfrac{2}{6}< \dfrac{3}{6}< \dfrac{4}{6}\)

hay \(\dfrac{1}{3}< \dfrac{1}{2}< \dfrac{2}{3}\)

b) Ta có:

+) \(\dfrac{4}{9}=\dfrac{56}{126}\)

+) \(-\dfrac{1}{2}=-\dfrac{63}{126}\)

+) \(\dfrac{3}{7}=\dfrac{54}{126}\)

=> \(-\dfrac{63}{126}< \dfrac{54}{126}< \dfrac{56}{126}\)

hay \(-\dfrac{1}{2}< \dfrac{3}{7}< \dfrac{4}{9}\)

c) Ta có:

+) \(\dfrac{27}{82}=\dfrac{2025}{6150}\)

+) \(\dfrac{26}{75}=\dfrac{2132}{6150}\)

=> \(\dfrac{2025}{6150}< \dfrac{2132}{6150}\)

hay \(\dfrac{27}{82}< \dfrac{26}{75}\)

d) Ta có:

+) \(-\dfrac{49}{78}=-\dfrac{4655}{7410}\)

+) \(-\dfrac{64}{95}=-\dfrac{4992}{7410}\)

=> \(-\dfrac{4665}{7410}>-\dfrac{4992}{7410}\)

hay \(-\dfrac{49}{78}>-\dfrac{64}{95}\)

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