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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 \(\frac{1}{5^9}<\frac{1}{3^9};\frac{1}{5^8}<\frac{1}{3^8};....;\frac{1}{5}<\frac{1}{3}\)nên \(\frac{1+5+5^2+...+5^8}{5^9}<\frac{1+3+3^2+...+3^8}{3^9}\)
=> \(\frac{1}{\frac{1+5+5^2+...+5^8}{5^9}}>\frac{1}{\frac{1+3+3^2+...+3^8}{3^9}}\)=> 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).
Ta có:
\(4\left(1+5+5^2+...+5^9\right)=5\left(1+5+5^2+...+5^9\right)-\left(1+5+5^2+...+5^9\right)\)
\(=5+5^2+5^3+...+5^{10}-1-5-5^2-...-5^9\)
\(=5^{10}-1+\left(5-5\right)+\left(5^2-5^5\right)+..+\left(5^9-5^9\right)\)
\(=5^{10}-1\)
=> \(1+5+5^2+...+5^9=\frac{5^{10}-1}{4}\)
Tương tự: \(1+5+5^2+...+5^8=\frac{5^9-1}{4}\)
\(1+3+3^2+...+3^9=\frac{3^{10}-1}{2}\)
\(1+3+3^2+...+3^8=\frac{3^9-1}{2}\)
=> \(A=\frac{5^{10}-1}{5^9-1}>\frac{5^{10}-1}{5^9}=5-\frac{1}{5^9}>4;\)
\(B=\frac{3^{10}-1}{3^9-1}< \frac{3^{10}}{3^9-1}=3+\frac{3}{3^9-1}< 4;\)
=> A > B.