Question

[Công chúa Phương Thìn]

\(CHO:\)

\(A=\frac{1+5+5^2+5^3+....+5^9}{1+5+5^2+5^3+....+5^8}\)

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

\(CMR:A>B\)

Answer 1

[o0o I am a studious person o0o]

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

Answer 2

[o0o I am a studious person o0o]

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 \(\Rightarrow A>B\left(đpcm\right)\)