Ta có: \(C=\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+...+\frac{1}{51.52}\)C bé hơn\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{50.52}=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{50}-\frac{1}{52}\right)\)

C bé hơn \(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{52}\right)\)bé hơn\(\frac{1}{2}.\frac{1}{2}=\frac{1}{4}\)(đpcm)

xin lỗi nha mk ko biết viết kí hiệu bé hơn

mik cảm ơn nha

\(\left(1-\frac{2}{42}\right)\left(1-\frac{2}{56}\right)...\left(1-\frac{2}{2652}\right)\)

= \(\frac{40}{42}.\frac{54}{56}.\frac{70}{72}...\frac{2650}{2652}\)

= \(\frac{5.8}{6.7}.\frac{6.9}{7.8}.\frac{7.10}{8.9}...\frac{50.53}{51.52}\)

= \(\frac{5.6.7...50}{6.7.8...51}.\frac{8.9.10...53}{7.8.9...52}\)

= \(\frac{5}{51}.\frac{53}{7}=\frac{265}{357}\)

A = 1/20 + 1/72 = 23/360

B = 1/2 + 1/6 + 1/30 = 7/10

C = 1/42 + 1/56 + 1/12 = 1/8

Trung bình cộng của A, B và C là: (23/360 + 7/10 + 1/8) : 3 = 8/27

Đáp số: 8/27

\(C=\frac{8}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(=\frac{8}{90}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)

\(=\frac{8}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)

\(=\frac{4}{45}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)

\(=\frac{4}{45}-\left(1-\frac{1}{9}\right)=\frac{4}{45}-\frac{8}{9}=\frac{4}{45}-\frac{40}{45}=\frac{-36}{45}=\frac{-4}{5}\)

\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)

\(=\dfrac{9}{10}-\left(\dfrac{1}{90}+\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\right)\)

\(=\dfrac{9}{10}-\left(\dfrac{1}{9\cdot10}+\dfrac{1}{9\cdot8}+\dfrac{1}{7\cdot8}+\dfrac{1}{7\cdot6}+\dfrac{1}{5\cdot6}-\dfrac{1}{5\cdot4}-\dfrac{1}{3\cdot4}-\dfrac{1}{3\cdot2}-\dfrac{1}{1\cdot2}\right)\)

\(=\dfrac{9}{10}-\left(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{8}-\dfrac{1}{9}+...+1-\dfrac{1}{2}\right)\)

\(=\dfrac{9}{10}-\left(1-\dfrac{1}{10}\right)\)

\(=\dfrac{9}{10}-\dfrac{9}{10}\)

\(=0\)