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\(\left(1-\frac{1}{3}\right)\)\(x\left(1-\frac{1}{6}\right)\)\(x\)\(\left(1-\frac{1}{10}\right)\)\(x\)\(\left(1-\frac{1}{15}\right)\)\(x\)\(\left(1-\frac{1}{21}\right)\)\(x\)\(\left(1-\frac{1}{28}\right)\)\(=\)\(\left(\frac{3}{3}-\frac{1}{3}\right)\)\(x\)\(\left(\frac{6}{6}-\frac{1}{6}\right)\)\(x\)\(\left(\frac{10}{10}-\frac{1}{10}\right)\)\(x\)\(\left(\frac{15}{15}-\frac{1}{15}\right)\)\(x\)\(\left(\frac{21}{21}-\frac{1}{21}\right)\)\(x\)\(\left(\frac{28}{28}-\frac{1}{28}\right)\)\(=\)\(\frac{2}{3}x\frac{5}{6}x\frac{9}{10}x\frac{14}{15}x\frac{20}{21}x\frac{27}{28}\)\(=\)\(\frac{2x5x9x14x20x27}{3x6x10x15x21x28}\)\(=\)\(\frac{2x5\left(3x3\right)x\left(2x7\right)x\left(5x4\right)x\left(3x3x3\right)}{3x\left(3x2\right)x\left(5x2\right)x\left(5x3\right)x\left(7x3\right)x\left(4x7\right)}\)\(=\)\(\frac{3}{7}\)
\(D=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\)
\(D=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+\frac{2}{90}\)
\(D=\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+\frac{2}{9.10}\)
\(D=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(D=2\left(\frac{1}{4}-\frac{1}{10}\right)=2\cdot\frac{3}{20}=\frac{3}{10}\)
\(E=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(E=\frac{5}{28}+\frac{1}{14}+\frac{1}{26}+...+\frac{1}{140}\)
\(E=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(E=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{25.28}\)
\(E=\frac{5}{3}\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(E=\frac{5}{3}\cdot\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}\cdot\frac{3}{14}=\frac{5}{14}\)
\(S=\frac{1}{3}+\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{300}\)
\(\Rightarrow\frac{1}{2}S=\frac{1}{6}+\frac{1}{12}+\cdot\cdot\cdot+\frac{1}{600}\)
\(\Rightarrow\frac{1}{2}S=\frac{1}{2\times3}+\frac{1}{3\times4}+\cdot\cdot\cdot+\frac{1}{24\times25}\)
\(\Rightarrow\frac{1}{2}S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdot\cdot\cdot+\frac{1}{24}-\frac{1}{25}\)
\(\Rightarrow\frac{1}{2}S=\frac{1}{2}-\frac{1}{25}\)
\(\Rightarrow\frac{1}{2}S=\frac{23}{50}\)
\(\Rightarrow S=\frac{23}{50}:\frac{1}{2}\)
\(\Rightarrow S=\frac{23}{25}\)
S = \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{300}\)
= \(2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{600}\right)\)
= \(2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{24\times25}\right)\)
= \(2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{24}-\frac{1}{25}\right)\)
= \(2\times\left(\frac{1}{2}-\frac{1}{25}\right)\)
\(=2\times\frac{23}{50}\)
\(=\frac{23}{25}\)
Bỏ đi \(\frac{1}{36}\) ta được
\(\frac{1}{3}\) +\(\frac{1}{6}\)+\(\frac{1}{10}\)+\(\frac{1}{15}\)+\(\frac{1}{21}\)+\(\frac{1}{28}\)
=\(\frac{140+70+42+28+20+15}{420}\)
=\(\frac{315}{420}\)
=\(\frac{3}{4}\)
Ta có : \(\frac{1}{4}+\frac{1}{3}:\frac{1}{x}=\frac{11}{12}\)
\(\Rightarrow\frac{1}{3}:\frac{1}{x}=\frac{11}{12}-\frac{1}{4}\)
\(\frac{1}{3}:\frac{1}{x}=\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{3}:\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{3}\times\frac{3}{2}\)
\(\frac{1}{x}=\frac{1}{2}\)
=> x = 2
a) \(\frac{x\div3-16}{2}+21=38\)
\(\frac{x\div3-16}{2}=38+21\)
\(\frac{x\div3-16}{2}=59\)
\(x\div3-16=59.2\)
\(x\div3-16=118\)
\(x\div3=118+16\)
\(x\div3=134\)
\(x=134.3\)
\(x=402\)
b) \(\frac{1}{4}+\frac{1}{3}\div\frac{1}{x}=\frac{11}{12}\)
\(\frac{1}{3}\div\frac{1}{x}=\frac{11}{12}-\frac{1}{4}\)
\(\frac{1}{3}\div\frac{1}{x}=\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{3}\div\frac{2}{3}\)
\(\frac{1}{x}=\frac{1}{2}\)
Vậy x = ....
Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}\)
\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{7}-\frac{1}{8}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{8}\right)\)
\(\Rightarrow A=2\cdot\frac{3}{8}=\frac{3}{4}\)