a) Đặt A = \(\frac{5^{12}+1}{5^{13}+1}\Rightarrow5A=\frac{5^{13}+5}{5^{13}+1}=1+\frac{4}{5^{13}+1}\)
Đặt \(B=\frac{5^{11}+1}{5^{12}+1}\Rightarrow5B=\frac{5^{12}+5}{5^{12}+1}=1+\frac{4}{5^{12}+1}\)
Vì \(\frac{4}{5^{13}+1}< \frac{4}{5^{12}+1}\Rightarrow1+\frac{4}{5^{13}+1}< 1+\frac{4}{5^{12}+1}\Rightarrow5A< 5B\Rightarrow A< B\)
Áp dụng công thức : \(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\left(a;b;m\in N\right)\)
Ta có : \(A=\frac{5^{12}+1}{5^{13}+1}< 1\)
\(\Leftrightarrow A=\frac{5^{12}+1}{5^{13}+1}< \frac{5^{12}+1+4}{5^{13}+1+4}=\frac{5^{12}+5}{5^{13}+5}=\frac{5\left(5^{11}+1\right)}{5\left(5^{12}+1\right)}=B\)
\(\Leftrightarrow A< B\)
b ) Đặt \(A=\frac{7^{10}}{1+7+7^2+...+7^9}\) nên \(\frac{1}{A}=\frac{1+7+7^2+...+7^9}{7^{10}}\)
\(=\frac{1}{7^{10}}+\frac{7}{7^{10}}+\frac{7^2}{7^{10}}+...+\frac{7^9}{7^{10}}=\frac{1}{7^{10}}+\frac{1}{7^9}+\frac{1}{7^8}+...+\frac{1}{7}\)
Đặt \(B=\frac{5^{10}}{1+5+5^2+...+5^9}\)nên \(\frac{1}{B}=\frac{1+5+5^2+...+5^9}{5^{10}}\)
\(=\frac{1}{5^{10}}+\frac{5}{5^{10}}+\frac{5^2}{5^{10}}+...+\frac{5^9}{5^{10}}=\frac{1}{5^{10}}+\frac{1}{5^9}+\frac{1}{5^8}+...+\frac{1}{5}\)
Ta thấy : \(\frac{1}{7^{10}}< \frac{1}{5^{10}};\frac{1}{7^9}< \frac{1}{5^9};...;\frac{1}{7}< \frac{1}{5}\)nên \(\frac{1}{A}< \frac{1}{B}\)
Vậy \(A< B\)
