\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}.\)

\(=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}....+\frac{2}{48.50}\right)\)

\(=\frac{1}{2}.\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{50-48}{48.50}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+.....+\frac{1}{48}-\frac{1}{50}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)

\(=\frac{1}{2}.\frac{12}{25}=\frac{6}{25}\)

\(B=\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{97.100}\)

\(=\frac{4-1}{1.4}+\frac{7-4}{4.7}+....+\frac{100-97}{97.100}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

\(C=\frac{8}{7.14}+\frac{8}{14.21}+....+\frac{8}{91.98}\)

\(=\frac{7}{8}.\left(\frac{7}{7.14}+\frac{7}{14.21}+...+\frac{7}{91.98}\right)\)

\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{14}+\frac{1}{14}-\frac{1}{21}+.....+\frac{1}{91}-\frac{1}{98}\right)\)

\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{98}\right)\)

\(=\frac{7}{8}.\frac{13}{98}=\frac{13}{112}\)

\(\frac{1.2+3.6+5.10+7.14}{2.3+6.9+10.15+14.21}\)

\(=\frac{1.2+3.6+5.10+7.14}{1.2.3+3.6.3+5.10.3+7.14.3}\)

\(=\frac{1.2+3.6+5.10+7.14}{3.\left(1.2+3.6+5.10+7.14\right)}\)

\(=\frac{1}{3}\)

NHAN A VOI 3 RUI TU TINH

DỄ MÀ

4A=\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{107.111}\)

4A=\(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{107}-\frac{1}{111}\)

4A=\(\frac{1}{3}-\frac{1}{111}=\frac{12}{37}\)

A=\(\frac{12}{37}:4=\frac{12}{37}.\frac{1}{4}=\frac{3}{37}\)

\(4A=\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{107.111}\)

\(4A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{107}-\frac{1}{111}\)

\(4A=\frac{1}{3}-\frac{1}{111}\)

\(4A=\frac{36}{111}\)

\(A=\frac{36}{111}\div4\)

\(A=\frac{9}{111}\)

\(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-...-\frac{1}{23.27}=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{23.27}\right)=\frac{1}{2}-\frac{1}{4}\cdot\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\right)=\frac{1}{2}-\frac{1}{4}\cdot\left(\frac{1}{3}-\frac{1}{27}\right)=\frac{1}{2}-\frac{1}{4}\cdot\frac{8}{27}=\frac{23}{54}\)

làm hộ mình vs khẩn cấp

A=1/3*7+1/7*11+..+1/95*99

=> 4A=4/3*7+4/7*11+..+4/95*99

=>4A=1/3-1/7+1/7-1/11+...+1/95-1/99=1/3-1/99=32/99

=>A=8/99

\(=\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+.......+\frac{4}{95.99}\right)=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=\frac{1}{4}.\frac{32}{99}=\frac{8}{99}\)

\(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{95.99}\)

\(=\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{95.99}\right)\)

\(=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\right)\)

\(=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=\frac{1}{4}\cdot\frac{32}{99}=\frac{8}{99}\)

Ta có A = \(\frac{4}{3.7}+\frac{4}{7.11}+..............+\frac{4}{107.111}\)

=> A = \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.............+\frac{1}{107}-\frac{1}{111}\)

A = \(\frac{1}{3}-\frac{1}{111}=\frac{12}{37}\)

k nha bạn