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\(E=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+...+\frac{1}{73\cdot75}\)
\(E=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)
\(\Rightarrow E=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)
\(F=\frac{15}{90\cdot94}+\frac{15}{94\cdot98}+...+\frac{15}{146\cdot150}\)
\(F=\frac{15}{4}\cdot\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)
\(\Rightarrow F=\frac{15}{4}\cdot\left(\frac{1}{90}-\frac{1}{150}\right)=\frac{15}{4}\cdot\frac{1}{225}=\frac{1}{60}\)
\(G=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(G=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(G=\frac{5}{4\cdot7}+\frac{5}{7\cdot10}+\frac{5}{10\cdot13}+...+\frac{5}{25\cdot28}\)
\(G=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(\Rightarrow G=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}\cdot\frac{3}{14}=\frac{5}{14}\)
Ta có :
\(H=\frac{15}{90.94}+\frac{15}{94.98}+\frac{15}{98.102}+...+\frac{15}{146.150}\)
\(H=\frac{15}{4}\left(\frac{4}{90.94}+\frac{4}{94.98}+\frac{4}{98.102}+...+\frac{4}{146.150}\right)\)
\(H=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+...+\frac{1}{146}-\frac{1}{150}\right)\)
\(H=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{150}\right)\)
\(H=\frac{15}{4}.\frac{1}{225}\)
\(H=\frac{1}{60}\)
Vậy \(H=\frac{1}{60}\)
Chúc bạn học tốt ~
\(H=\frac{15}{90\cdot94}+\frac{15}{94\cdot98}+\frac{15}{98\cdot102}+...+\frac{15}{146\cdot150}\)
\(H=15\left(\frac{1}{90\cdot94}+\frac{1}{94\cdot98}+\frac{1}{98\cdot102}+...+\frac{1}{146\cdot150}\right)\)
\(H=15\left[\frac{1}{4}\left(\frac{4}{90\cdot94}+\frac{4}{94\cdot98}+\frac{4}{98\cdot102}+...+\frac{4}{146\cdot150}\right)\right]\)
\(H=15\left[\frac{1}{4}\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+...+\frac{1}{146}-\frac{1}{150}\right)\right]\)
\(H=15\left[\frac{1}{4}\left(\frac{1}{90}-\frac{1}{150}\right)\right]\)
\(H=15\left[\frac{1}{4}\cdot\frac{1}{225}\right]\)
\(H=15\cdot\frac{1}{900}\)
\(H=\frac{1}{60}\)
a) \(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
\(A=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(A=2\cdot\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{15\cdot16}\right)\)
\(A=2\cdot\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(A=2\cdot\left(\frac{1}{4}-\frac{1}{16}\right)=2\cdot\frac{3}{16}=\frac{3}{8}\)
b) \(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(B=\frac{5}{3}\cdot\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{25\cdot28}\right)\)
\(B=\frac{5}{3}\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(B=\frac{5}{3}\cdot\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}\cdot\frac{3}{14}=\frac{5}{14}\)
\(A=\frac{15}{90.94}+\frac{15}{94.98}+...+\frac{15}{146.150}\)
\(A=\frac{4}{4}\left(\frac{15}{90.94}+\frac{15}{94.98}+...+\frac{15}{146.150}\right)\)
\(A=\frac{15}{4}\left(\frac{4}{90.94}+\frac{4}{94.98}+...+\frac{4}{146.150}\right)\)
\(A=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)
\(A=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{150}\right)\)
\(A=\frac{15}{4}.\frac{1}{225}=\frac{1}{60}\)
\(B=\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\)
\(B=\frac{3}{3}\left(\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\right)\)
\(B=2\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)
\(B=2\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(B=2\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(B=2.\frac{1}{18}=\frac{1}{9}\)
Trả lời:
\(A=\frac{15}{90.94}+\frac{15}{94.98}+...+\frac{15}{146.150}\)
\(A=\frac{15}{4}.\left(\frac{4}{90.94}+\frac{4}{94.98}+...+\frac{4}{146.150}\right)\)
\(A=\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)
\(A=\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{150}\right)\)
\(A=\frac{15}{4}.\frac{1}{225}\)
\(A=\frac{1}{60}\)
\(B=\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\)
\(B=2.\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)
\(B=2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(B=2.\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(B=2.\frac{1}{18}\)
\(B=\frac{1}{9}\)
\(A=\frac{15}{90.94}+\frac{15}{94.98}+...+\frac{15}{146.150}\)
\(A=15\left(\frac{1}{90.94}+\frac{1}{94.98}+...+\frac{1}{146.150}\right)\)
\(A=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)
\(A=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{150}\right)\)
\(A=\frac{15}{4}.\frac{1}{225}\)
\(A=\frac{1}{60}\)
\(\frac{15}{90.94}+\frac{15}{94.98}+\frac{15}{98.102}+...+\frac{15}{146.150}\)
\(=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+...+\frac{1}{146}-\frac{1}{150}\right)\)
\(=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{150}\right)\)
\(=\frac{15}{4}.\frac{1}{225}\)
\(=\frac{1}{60}\)
\(\frac{15}{90\cdot94}+\frac{15}{94\cdot98}+\frac{15}{98\cdot102}+...+\frac{15}{146\cdot150}\)
\(=\frac{15}{4}\cdot\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+...+\frac{1}{146}-\frac{1}{150}\right)\)
\(=\frac{15}{4}\cdot\left(\frac{1}{90}-\frac{1}{150}\right)\)
\(=\frac{15}{4}\cdot\frac{1}{225}=\frac{1}{60}\)
\(A=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)
=>\(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)
=>\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)
=>\(A=\frac{1}{3}-\frac{1}{21}\)
=>\(A=\frac{2}{7}\)
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