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Trả lời:

A = \(\frac{3^{10}+1}{3^9+1}=\frac{3.3^9+1}{3.3^8+1}=\frac{3^9+1}{3^8+1}\)= B

_Học tốt bạn nha_

A=1+5+5^2+..+5^9/1+5+5^2+...+5^8

=1+5^9/1+5+5^2+...+5^8 

B=1+3+3^2+..+3^9/1+3+3^2+..+3^8

=1+3^9/1+3+3^2+..+3^8

đặt A' =1+5+5^2+...+5^8

5A'=5+5^2+5^3+...+5^9

5A'-A'=5+5^2+5^3+...+5^9-5-1-5-5^2-...-5^8

4A'=5^9-1=>A'=(5^9-1):4

tương tự B'=(3^9-1):4

A=1+5^9/(5^9-1)/4=4.5^9/5^9-1

B=1+3^9/(3^9-1)/4=4.3^9/3^9-1

=> A<B

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

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

\(=5+\frac{1}{1+3+3^2+....+3^8}-2\)  

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

\(\Rightarrow A>B\)

mk nghĩ là 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 \(3^9+1>3^8+1\)

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

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

\(\Rightarrow A>B\)

2/5 x 1/X + 1/X x 2 = 0,1

1/X x ( 2/5 + 2 ) = 0,1

1/X x 12 / 5 = 0,1

1/X             = 0,1 :12/5 = 1/10 : 12/5

1/X             = 1/24

Vậy X = 24