Ta có: B > 1
=> B = \(\frac{2^{10}-1}{2^{10}-3}>\frac{2^{10}-1+2}{2^{10}-3+2}=\frac{2^{10}+1}{2^{10}-1}=A\)
Vậy A < B
\(\frac{2^{10}+1}{2^{10}-1}=\frac{2^{10}-1+2}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)
\(\frac{2^{10}-1}{2^{10}-3}=\frac{2^{10}-3+2}{2^{10}-3}=1+\frac{2}{2^{10}-3}\)
Nhận thấy: \(\frac{2}{2^{10}-3}>\frac{2}{2^{10}-1}\) do 210-3 < 210-1
Vậy: \(\frac{2^{10}-1}{2^{10}-3}>\frac{2^{10}+1}{2^{10}-1}\)
Ta có : \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{b+n}\)
Nên : \(B=\frac{2^{10}-1}{2^{10}-3}>\frac{2^{10}-1+2}{2^{10}-3+2}=\frac{2^{10}+1}{2^{10}-1}\)
Good