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A=1/1+5+5^2+5^3+...+5^8+5+5^2+5^3+...+5^9=1/1+5+5^2+5^3+...+5^8+5.
Tương tự B=1/1+3+3^2+...+3^8+3
=>A>B.
k nha.
\(A=\frac{1+5+5^2+...+5^8}{1+5+5^2+...+5^8}+\frac{5^9}{1+5+5^2+...+5^8}=1+\frac{5^9}{1+5+5^2+....+5^8}=1+\frac{1}{\frac{1+5+5^2+...+5^8}{5^9}}\)
\(B=\frac{1+3+3^2+...+3^8}{1+3+3^2+...+3^8}+\frac{3^9}{1+3+3^2+...+3^8}=1+\frac{1}{\frac{1+3+3^2+....+3^8}{3^9}}\)
Nhận xét:
\(\frac{1+5+5^2+...+5^8}{5^9}=\frac{1}{5^9}+\frac{1}{5^8}+\frac{1}{5^7}+...+\frac{1}{5}\); \(\frac{1+3+3^2+...+3^8}{3^9}=\frac{1}{3^9}+\frac{1}{3^8}+\frac{1}{3^7}+....+\frac{1}{3}\)
Vì \(\frac{1}{5^9}
\(A=1+\frac{5^9}{1+5+..+5^8}\)
\(=1+\frac{1}{\frac{1}{5^9}+\frac{1}{5^8}+...+\frac{1}{5}}\)
Tương tự:
\(B=1+\frac{1}{\frac{1}{3^9}+\frac{1}{3^8}+...+\frac{1}{3}}\)
Vì \(\frac{1}{5}< \frac{1}{3}\) , \(\frac{1}{5^2}< \frac{1}{3^2}\), . . .
nên: \(\frac{1}{\frac{1}{5^9}+\frac{1}{5^8}+...+\frac{1}{5}}>\frac{1}{\frac{1}{3^9}+\frac{1}{3^8}+...+\frac{1}{3}}\)
=> A > B
Vậy đề bạn cho chứng minh A < B là sai nhé.
Ta có:\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}\)
=>\(A=\frac{\left(1+5+5^2+...+5^8\right)}{\left(1+5+5^2+...+5^8\right)}+\frac{5^9}{1+5+5^2+...+5^8}\)
=>\(A=1+\frac{5^9}{1+5+5^2+...+5^8}\)
Ta có:\(B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}\)
=>\(B=\frac{1+3+3^2+...+3^8}{1+3+3^2+...+3^8}+\frac{3^9}{1+3+3^2+...+3^8}\)
=>\(B=1+\frac{3^9}{1+3+3^2+...+3^8}\)
vì:\(1+3+3^2+...+3^8< 1+5+5^2+...+5^8\)
Nên A<B(đpcm).
\(A=\frac{1+5+5^2+...+5^8+5^9}{1+5+5^2+...+5^8}=1+\frac{5^9}{5^8}=6\)
\(B=\frac{1+3+3^2+...+3^8+3^9}{1+3+3^2+...+3^8}=1+\frac{3^9}{3^8}=4\)
Từ đó suy ra A>B
kieu nay la ko tinh ra ket qua hay so sanh
A=1+C; voi C=5^9/(1+...5^8)=1/(1/5^9+1/5^8+...+1/5)
B=1+D;voi D=3^9/(1+..3^8)=1/(1/3^9+1/3^8+...+1/3)
C=1/E; voi E=(1/5^9+1/5^8+...+1/5)
D=1/f; voi F=(1/3^9+1/3^8+...+1/3)
=> F-E=(1/3-1/5)+...+(1/3^9-1/5^9) >0=> F>E
=> C>D=> 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 \(\frac{1}{1+3+3^2+....+3^8}-2< 0\)
\(\Rightarrow A>B\)
\(\frac{1}{5}A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^9}=1\)
\(\frac{1}{3}B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^9}=1\)
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
