Ta có: \(A=\frac{23}{1\cdot300}+\frac{23}{2\cdot301}+\frac{23}{3\cdot302}+\cdots+\frac{23}{101\cdot400}\)

\(=23\left(\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\cdots+\frac{1}{101\cdot400}\right)\)

\(=\frac{23}{299}\left(\frac{299}{1\cdot300}+\frac{299}{2\cdot301}+\cdots+\frac{299}{101\cdot400}\right)\)

\(=\frac{23}{299}\left(1-\frac{1}{300}+\frac12-\frac{1}{301}+\cdots+\frac{1}{101}-\frac{1}{400}\right)\)

\(=\frac{1}{13}\left(1+\frac12+\cdots+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-\cdots-\frac{1}{400}\right)\)

Ta có: \(B=\frac{1313}{1\cdot102}+\frac{1313}{2\cdot103}+\cdots+\frac{1313}{299\cdot400}\)

\(=13\left(\frac{101}{1\cdot102}+\frac{101}{2\cdot103}+\cdots+\frac{101}{299\cdot400}\right)\)

\(=13\left(1-\frac{1}{102}+\frac12-\frac{1}{103}+\cdots+\frac{1}{299}-\frac{1}{400}\right)\)

\(=13\left(1+\frac12+\cdots+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-\cdots-\frac{1}{400}\right)\)

Do đó; \(\frac{A}{B}=\frac{1}{13}:13=\frac{1}{169}\)

Bạn giải thích cho mình đoạn này được không ? Cảm ơn bạn .

\(\frac{4354}{23}+\frac{2453425}{23}=\frac{4354+2453425}{23}=\frac{2457779}{23}\)

\(\frac{43352}{12}+\frac{23412134}{23}=\frac{997096}{276}+\frac{280945608}{276}=\frac{281942704}{276}\)

\(\frac{4354+2453425}{23}=\frac{2457779}{23}\)

Bn hỏi cái gì vậy hả??/

mấy số đó là bao nhiêu ? nhưng cũng thank bạn nhiều vì nhắc cho mình

\(B=\frac{23^{41}+1}{23^{42}+1}\)

Vì B < 1

\(\Rightarrow B=\frac{23^{41}+1}{23^{42}+1}< \frac{23^{41}+1+22}{23^{42}+1+22}=\frac{23^{41}+23}{23^{42}+23}=\frac{23(23^{40}+1)}{23\left(23^{41}+1\right)}=\frac{23^{40}+1}{23^{41}+1}=A\)

P/s: Hoq chắc

ta có 

\(B=\frac{23^{41}+1}{23^{42}+1}< \frac{23^{41}+1+22}{23^{42}+1+22}=\frac{23^{41}+23}{23^{42}+23}=\frac{23\left(23^{40}+1\right)}{23\left(23^{41}+1\right)}=\frac{23^{40}+1}{23^{41}+1}=A\)

\(\Rightarrow B< A\)

=-11/23x(6/7+8/7)-1/23

=-11/23x2-1/23

=-22/23-1/23

=-1

\(\frac{-11}{23}.\frac{6}{7}+\frac{8}{7}.\frac{-11}{23}-\frac{1}{23}\)=\(\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)=\(\frac{-11}{23}.2-\frac{1}{23}\)=\(\frac{-22}{23}-\frac{1}{23}=\frac{-21}{23}\)

= -2/3

\(\frac{-29}{161}\)

\(\frac{-29}{161}\)

\(\dfrac{-5}{23}\cdot\dfrac{13}{17}+\dfrac{-5}{23}\cdot\dfrac{4}{17}+\dfrac{5}{23}\\ =\dfrac{-5}{23}\cdot\left(\dfrac{13}{17}+\dfrac{4}{17}\right)+\dfrac{5}{23}\\ =\dfrac{-5}{23}\cdot1+\dfrac{5}{23}\\ =\dfrac{-5}{23}+\dfrac{5}{23}\\ =0\)

\(=\frac{23^{10}\left(1-23\right)}{23^{10}\left(23+21\right)}=\frac{-22}{44}=\frac{-1}{2}\)

\(\frac{23^{10}-23^{11}}{23^{11}+21.23^{10}}\)

\(=\frac{23^{10}.\left(1-23\right)}{23^{10}.\left(23+21\right)}\)

\(=\frac{-22}{44}=\frac{-22}{22.2}=\frac{-1}{2}\)