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 \(\Rightarrow A>B\left(đpcm\right)\)
