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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; \(\frac{A}{B}=\frac{1}{13}:13=\frac{1}{169}\)
\(B=\frac{23^{41}+1}{23^{42}+1}\)
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
\(B=\frac{23^{41}+1}{23^{42}+1}< \frac{23^{41}+1+22}{23^{42}+1+22}=\frac{23^{41}+23}{23^{42}+23}=\frac{23\left(23^{40}+1\right)}{23\left(23^{41}+1\right)}=\frac{23^{40}+1}{23^{41}+1}=A\)
\(\Rightarrow B< A\)
\(\frac{-11}{23}.\frac{6}{7}+\frac{8}{7}.\frac{-11}{23}-\frac{1}{23}\)=\(\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)=\(\frac{-11}{23}.2-\frac{1}{23}\)=\(\frac{-22}{23}-\frac{1}{23}=\frac{-21}{23}\)
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Bạn kiểm tra lại đề nha.
sửa đề: \(-\frac{14}{23}-\frac{9}{23}\)
\(-\frac{14}{23}-\frac{9}{23}\)
\(=\frac{-14-9}{23}\)
\(=\frac{-23}{23}=-1\)