a=5^9

b=3^9

=>a>b

\(\frac{1}{5}A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^9}=1\)
\(\frac{1}{3}B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^9}=1\)
Vì  \(\frac{1}{5}<\frac{1}{3}\)Nên \(\frac{1}{5}A<\frac{1}{5}B\)
Vậy A<B

ai trả lời cũng sai hết rồi 

Tui Gợi ý là A > B

Bây giờ các bạn ghi cách giải đi

sao hong ai dê y vay troi

mình viết tắt bạn tự hiểu nha:

a=1+(5⁹/1+5+5²5+...+5⁸

b=1+(3⁹/1+3+3³+....+3⁸

VD:A/B-C/D=A.C/B.D-C.B/D.B

TƯƠNG TỰ NHƯ A,B BẠN TÍNH RA

Ta có: \(5\left(1+5+5^2+...+5^9\right)-\left(1+5+5^2+...+5^9\right)\)

= \(\left(5+5^2+5^3+...+5^{10}\right)-\left(1+5+5^2+...+5^9\right)\)

\(4\left(1+5+5^2+...+5^9\right)\)\(=5^{10}-1\)

=> \(1+5+5^2+...+5^9=\frac{5^{10}-1}{4}\)

Tương tự: \(1+5+5^2+....+5^8=\frac{5^9-1}{4}\)

=> \(A=\frac{\frac{5^{10}-1}{4}}{\frac{5^9-1}{4}}=\frac{5^{10}-1}{5^9-1}=\frac{5\left(5^9-1\right)+4}{5^9-1}=5+\frac{4}{5^9-1}>5\)

Tương tự:

\(1+3+3^2+...+3^9=\frac{3^{10}-1}{2}\)

và \(1+3+3^2+...+3^8=\frac{3^9-1}{2}\)

=>\(B=\frac{3^{10}-1}{3^9-1}=\frac{3\left(3^9-1\right)+2}{3^9-1}=3+\frac{2}{3^9-1}< 5\)

=> A >  5 > B

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

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

mà \(\frac{1}{1+5+5^2+...+5^8}\approx0\)

suy ra: A= 5.

chứng minh tương tự, ta có: B=3

5 > 3 --> 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