\(\frac{3}{12}+\frac{3}{20}+\frac{3}{30}+...+\frac{3}{110}\)

\(=3\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{110}\right)\)

\(=3\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{10\cdot11}\right)\)

\(=3\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{11}\right)\)

\(=3\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=3\cdot\frac{8}{33}=\frac{8}{11}\)

https://olm.vn/hoi-dap/detail/63463423750.html

vào tham khảo đi mk nhác trình bày

= 3.(1/12+1/20+1/30+1/42+1/56+1/72+1/90+1/110)

= 3.8/33

= 8/11

rút gọn thành

1/2 x 2/3 x 3/4 x 4/5 x 5/6 x 6/7 x 7/8 x 8/9

1x2x3x4x5x6x7x8

__________________

2x3x4x5x6x7x8x9

gạch các số giống nhau ở tử và mẫu và bằng 1/9

Giải:

\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-...-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{1}{1.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-...-\dfrac{1}{9.10}-\dfrac{1}{10.11}=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{-1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\right)=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{-1}{2}\left(\dfrac{1}{1}-\dfrac{1}{11}\right)=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{-1}{2}.\dfrac{10}{11}=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{-5}{11}=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{-5}{11}+\dfrac{5}{13}=x\)

\(\Leftrightarrow x=\dfrac{-10}{143}\)

Vậy ...

Giải đẳng cấp nhỉ @Hắc Hường

Ta có:

\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-...-\dfrac{1}{110}=x-\dfrac{5}{13}\\ -\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-...-\dfrac{1}{110}=x-\dfrac{5}{13}-\dfrac{1}{3}\\ -\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-...-\dfrac{1}{110}=x-\dfrac{28}{39}\\ -\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{110}\right)=x-\dfrac{28}{39}\\ -\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{10.11}\right)=x-\dfrac{28}{39}\\ -\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{10}-\dfrac{1}{11}\right)=x-\dfrac{28}{39}\\ -\left(\dfrac{1}{3}-\dfrac{1}{11}\right)=x-\dfrac{28}{39}\\ -\dfrac{8}{33}=x-\dfrac{28}{39}\\ x=-\dfrac{8}{33}+\dfrac{28}{39}\\ x=\dfrac{68}{143}\)

Vậy \(x=\dfrac{68}{143}\)

A - B = \(\left(1+\frac{1}{2}+1+\frac{1}{12}+1+\frac{1}{30}+1+\frac{1}{56}+1+\frac{1}{90}\right)-\left(1-\frac{1}{6}+1-\frac{1}{20}+1-\frac{1}{42}+1-\frac{1}{72}+1-\frac{1}{110}\right)\)= \(\left(5+\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+\frac{1}{90}\right)-\left(5-\frac{1}{6}-\frac{1}{20}-\frac{1}{42}-\frac{1}{72}-\frac{1}{110}\right)\)\

= \(5+\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+\frac{1}{90}-5+\frac{1}{6}+\frac{1}{20}+\frac{1}{42}+\frac{1}{72}+\frac{1}{110}\)

= \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}+\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{10.11}\)

= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}=1-\frac{1}{11}=\frac{10}{11}\)

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}+\frac{109}{110}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}+1-\frac{1}{110}\)

\(=10-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\right)\)

\(=10-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\right)\)

\(=10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\right)\)

\(=10-\left(1-\frac{1}{10}\right)\)

\(=\frac{91}{10}\)

\(-\frac{3}{20}+\frac{-3}{30}+\frac{-3}{42}+\frac{-3}{56}+\frac{-3}{72}+\frac{-3}{90}\)

\(=-3\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

\(=-3\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(=-3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=-3\left(\frac{1}{4}-\frac{1}{10}\right)=-\frac{9}{20}\)

\(\frac{-3}{20}+\frac{-3}{30}+\frac{-3}{42}+\frac{-3}{56}+\frac{-3}{72}+\frac{-3}{90}\)

\(=-3\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

\(=-3\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{9\cdot10}\right)\)

\(=-3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=-3\left(\frac{1}{4}-\frac{1}{10}\right)\)

\(=-3\cdot\frac{3}{20}\)

\(=\frac{-9}{20}\)

\(\frac{1}{3}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}-\frac{1}{90}-\frac{1}{110}=x-\frac{5}{13}\)

\(\frac{1}{3}-\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\right)=x-\frac{5}{13}\)

\(\frac{1}{3}-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\right)=x-\frac{5}{13}\)

\(\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\right)=x-\frac{5}{13}\)

\(\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{11}\right)=x-\frac{5}{13}\)

\(\frac{1}{3}-\frac{1}{3}+\frac{1}{11}=x-\frac{5}{13}\)

\(\frac{1}{11}=x-\frac{5}{13}\)

\(x=\frac{1}{11}+\frac{5}{13}\)

\(x=\frac{68}{143}\)