áp dụng tc \(\frac{a}{b}< 1\Rightarrow\frac{a+m}{a+m}< 1\left(m\in N\right)\)
Ta có: \(B=\frac{15^{16}+1}{15^{17}+1}< \frac{15^{16}+1+14}{15^{17}+1+14}\)\(=\frac{15^{16}+15}{15^{17}+15}=\frac{15.\left(15^{15}+1\right)}{15.\left(15^{16}+1\right)}=\frac{15^{15}+1}{15^{16}+1}\)
\(\Rightarrow B< A\)
\(A=\frac{15^{15}+1}{15^{16}+1}\)
\(\Rightarrow15A=\frac{15^{16}+15}{15^{16}+1}\)
\(\Rightarrow15A=\frac{15^{16}+1+14}{15^{16}+1}\)
\(\Rightarrow15A=\frac{15^{16}+1}{15^{16}+1}+\frac{14}{15^{16}+1}\)
\(\Rightarrow15A=1+\frac{14}{15^{16}+1}\)
\(B=\frac{15^{16}+1}{15^{17}+1}\)
\(\Rightarrow15B=\frac{15^{17}+15}{15^{17}+1}\)
\(\Rightarrow15B=\frac{15^{17}+1+14}{15^{17}+1}\)
\(\Rightarrow15B=\frac{15^{17}+1}{15^{17}+1}+\frac{14}{15^{17}+1}\)
\(\Rightarrow15B=1+\frac{14}{15^{17}+1}\)
Vì \(\frac{14}{15^{17}+1}< \frac{14}{15^{16}+1}\) nên \(15B< 15A\)
Vậy B < A
