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\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}=\frac{1+5\left(1 +5+5^2+...+5^8\right)}{1+5+5^2+...+5^8}=5+\frac{1}{1+5+5^2+...+5^8} \)
\(B=\frac{1+3+3^2+....+3^9}{1+3+3^2+....+3^8}=\frac{1+3\left(1+3+3^2+....+3^8\right)}{1+3+3^2+....+3^8}=3+\frac{1}{1+3+3^2+....+3^8}\)
\(=5+\frac{1}{1+3+3^2+....+3^8}-2\)
Có: \(\frac{1}{1+5+5^2+...+5^8}>0\) và \(\frac{1}{1+3+3^2+....+3^8}-2< 0\)
\(\Rightarrow A>B\)
\(A=1+\frac{5^9}{1+5+..+5^8}\)
\(=1+\frac{1}{\frac{1}{5^9}+\frac{1}{5^8}+...+\frac{1}{5}}\)
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=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.
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
\(A=\frac{1+5+5^2+...+5^8}{1+5+5^2+...+5^8}+\frac{5^9}{1+5+5^2+...+5^8}=1+\frac{5^9}{1+5+5^2+....+5^8}=1+\frac{1}{\frac{1+5+5^2+...+5^8}{5^9}}\)
\(B=\frac{1+3+3^2+...+3^8}{1+3+3^2+...+3^8}+\frac{3^9}{1+3+3^2+...+3^8}=1+\frac{1}{\frac{1+3+3^2+....+3^8}{3^9}}\)
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}
Gọi tử là : R
=> \(R=1+5+5^2+5^3+......+5^9\)
\(\Rightarrow5R=5+5^2+5^3+....5^{10}\)
\(\Rightarrow5R-R=5^{10}-1\)
\(\Rightarrow4R=5^{10}-1\)
\(\Rightarrow R=\frac{5^{10}-1}{4}\)
Goij mẫu là M
\(\Rightarrow M=1+5+5^2+5^3+.....+5^8\)
\(\Rightarrow5M=5+5^2+.....+5^9\)
\(\Rightarrow5M-M=5^9-1\)
\(\Rightarrow M=\frac{5^9-1}{4}\)
\(\Rightarrow A=\frac{\frac{5^{10}-1}{4}}{\frac{5^9-1}{4}}=1\)
Tương tự : B
Rồi so sánh thôi dễ mà
Phần B nek :
Gọi tử là : T
\(\Rightarrow T=1+3+3^2+.....+3^9\)
\(\Rightarrow3T=3+3^2+3^3+.....+3^{10}\)
\(\Rightarrow3T-T=3^{10}-1\)
\(\Rightarrow T=\frac{3^{10}-1}{2}\)
Gọi mẫu là : H
\(\Rightarrow H=1+3+3^2+.....+3^8\)
\(\Rightarrow3H=3+3^2+3^3+.....+3^9\)
\(\Rightarrow3H-H=3^9-1\)
\(\Rightarrow H=\frac{3^9-1}{2}\)
\(\Rightarrow B=\frac{T}{H}=\frac{\frac{3^{10}-1}{2}}{\frac{3^9-1}{2}}=\frac{29524}{9841}=3,0001.....\)
Cho a sửa câu a nha :
\(\Rightarrow A=\frac{R}{M}=\frac{\frac{5^{10}-1}{4}}{\frac{5^9-1}{4}}=\frac{2441406}{488281}=5,000002048\)
Vậy \(\Rightarrow A>B\left(đpcm\right)\)