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

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

nhân 13 với A và B

so sánh mẫu =>mẫu nào lớn hơn thì bé hơn

là xong

em cần đáp án thui ạ, em cảm ơn nhiều lắm ạ

> nhé

\(A=\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{25}{156}\)

\(B=\dfrac{1}{10}+\dfrac{1}{15}=\dfrac{1}{6}\)

Quy đồng A và B ta được :

\(\dfrac{25}{156};\dfrac{26}{256}\)

=> A < B

\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{60}=\left(\frac{1}{11}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+...+\frac{1}{60}\right)_{\left(1\right)}>\frac{1}{20}.10+\frac{1}{30}.10+\frac{1}{40}.10+\frac{1}{50}.10+\frac{1}{60}.10=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{29}{20}=\frac{87}{60}>\frac{70}{60}=\frac{7}{6}=B\)

(1): mỗi nhóm có 10 số hạng

=>A>B