A<B

\(ta có A=\dfrac{13^{15}+1}{13^{16}+1}=\dfrac{13^{15}}{13^{16}}+1\)=\(\dfrac{1}{13}+1\)

B=\(\dfrac{13^{16}+1}{13^{17}+1}=\dfrac{13^{16}}{13^{17}}+1\)=\(\dfrac{1}{13}+1\)

vậy A=B

\(A=\dfrac{13^{15}+1}{13^{16}+1}vàB=\dfrac{13^{16}+1}{13^{17}+1}\)

ta có

\(\dfrac{13^{16}+1}{13^{17}+1}< 1\Rightarrow\dfrac{13^{16}+1+12}{13^{17}+1+12}=\dfrac{13\left(13^{15}+1\right)}{13\left(13^{16}+1\right)}=\dfrac{13^{15}+1}{13^{16}+1}=A\)

vậy B<A

\(A=\dfrac{13^{15}+1}{13^{16}+1}vàB=\dfrac{13^{16}+1}{13^{17}+1}\)

ta có B<1 nên

\(\dfrac{13^{16}+1}{13^{17}+1}< \dfrac{13^{16}+1+12}{13^{17}+1+12}=\dfrac{13\left(13^{15}+1\right)}{13\left(13^{16}+1\right)}=\dfrac{13^{15}+1}{13^{16}+1}=A\)

Vậy B<A

\(A=\dfrac{14^{14}+1}{14^{15}+1}\)

\(\Rightarrow14.A=\dfrac{14^{15}+14}{14^{15}+1}\)

\(\Rightarrow14.A=\dfrac{14^{15}+1}{14^{15}+1}+\dfrac{13}{14^{15}+1}\)

\(\Rightarrow14.A=1+\dfrac{13}{14^{15}+1}\)

\(B=\dfrac{14^{15}+1}{14^{16}+1}\)

\(\Rightarrow14.B=\dfrac{14^{16}+14}{14^{16}+1}\)

\(\Rightarrow14.B=\dfrac{14^{16}+1}{14^{16}+1}+\dfrac{13}{14^{16}+1}\)

\(\Rightarrow14.B=1+\dfrac{13}{14^{16}+1}\)

Nhận xét: \(\dfrac{13}{14^{15}+1}>\dfrac{13}{14^{16}+1}\) (cùng tử, xét mẫu)

\(\Rightarrow A>B\)

Vậy \(A>B\)

Ta có công thức : 

\(\frac{a}{b}< 1\) \(\Rightarrow\) \(\frac{a}{b}< \frac{a+c}{b+c}\)

\(\Rightarrow\)\(B=\frac{15^{16}+1}{15^{17}+1}< \frac{15^{16}+1+14}{15^{17}+1+14}=\frac{15^{16}+15}{15^{17}+15}=\frac{15\left(15^{15}+1\right)}{15\left(15^{16}+1\right)}=\frac{15^{15}+1}{15^{16}+1}=A\)

Vậy \(A>B\)

tại sao a/b<1 thì a/b<a+c/b+C

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

Ta có công thức \(\frac{a}{b}