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\(\frac{1.2+3.6+5.10+7.14}{2.3+6.9+10.15+14.21}\)
\(=\frac{1.2+3.6+5.10+7.14}{1.2.3+3.6.3+5.10.3+7.14.3}\)
\(=\frac{1.2+3.6+5.10+7.14}{3.\left(1.2+3.6+5.10+7.14\right)}\)
\(=\frac{1}{3}\)
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
\(=\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+.......+\frac{4}{95.99}\right)=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=\frac{1}{4}.\frac{32}{99}=\frac{8}{99}\)
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=\(\frac{12}{37}:4=\frac{12}{37}.\frac{1}{4}=\frac{3}{37}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{2}-\frac{1}{50}\)
\(=\frac{12}{25}\)
\(B=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{23.27}\)
\(=\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{23.27}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)\)
\(=\frac{1}{4}.\frac{8}{27}=\frac{2}{27}\)
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
\(a,\frac{1}{6}-\frac{2}{3}\)
\(=\frac{1}{6}-\frac{4}{6}=-\frac{3}{6}=-\frac{1}{2}\)
A =\(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{14}+\frac{1}{14}-\frac{1}{21}+...+\frac{1}{140}-\frac{1}{150}\)
A=\(\left(\frac{1}{3}-\frac{1}{150}\right)\)
A=\(\left(\frac{50}{150}-\frac{1}{150}\right)\)
A=\(\frac{49}{150}\)
\(A=\frac{1}{3x7}+\frac{1}{7x14}+....+\frac{1}{140x150}\)
\(A=\frac{1}{3}-\frac{1}{7}+.....+\frac{1}{140}-\frac{1}{150}\)
\(A=\frac{1}{3}-\frac{1}{150}\)
\(A=\frac{49}{150}\)