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\(\left(1-\frac{1}{3}\right)\)\(x\left(1-\frac{1}{6}\right)\)\(x\)\(\left(1-\frac{1}{10}\right)\)\(x\)\(\left(1-\frac{1}{15}\right)\)\(x\)\(\left(1-\frac{1}{21}\right)\)\(x\)\(\left(1-\frac{1}{28}\right)\)\(=\)\(\left(\frac{3}{3}-\frac{1}{3}\right)\)\(x\)\(\left(\frac{6}{6}-\frac{1}{6}\right)\)\(x\)\(\left(\frac{10}{10}-\frac{1}{10}\right)\)\(x\)\(\left(\frac{15}{15}-\frac{1}{15}\right)\)\(x\)\(\left(\frac{21}{21}-\frac{1}{21}\right)\)\(x\)\(\left(\frac{28}{28}-\frac{1}{28}\right)\)\(=\)\(\frac{2}{3}x\frac{5}{6}x\frac{9}{10}x\frac{14}{15}x\frac{20}{21}x\frac{27}{28}\)\(=\)\(\frac{2x5x9x14x20x27}{3x6x10x15x21x28}\)\(=\)\(\frac{2x5\left(3x3\right)x\left(2x7\right)x\left(5x4\right)x\left(3x3x3\right)}{3x\left(3x2\right)x\left(5x2\right)x\left(5x3\right)x\left(7x3\right)x\left(4x7\right)}\)\(=\)\(\frac{3}{7}\)
\(x\)là dấu nhân hả bạn? Nếu vậy thì mk làm cho nhé
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{20}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.......\cdot\frac{17}{18}\cdot\frac{18}{19}\cdot\frac{19}{20}=\frac{1}{20}\)
Vậy \(A=\frac{1}{20}\)
\(B=1\frac{1}{2}\cdot1\frac{1}{3}\cdot1\frac{1}{4}\cdot........\cdot1\frac{1}{2005}\cdot1\frac{1}{2006}\cdot1\frac{1}{2007}\)
\(B=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot......\cdot\frac{2006}{2005}\cdot\frac{2007}{2006}\cdot\frac{2008}{2007}=\frac{2008}{2}=1004\)
Vậy \(B=1004\)
DẤU CHẤM LÀ DẤU NHÂN
a,
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{19}{20}=\frac{1}{20}\)
b, \(1\frac{1}{2}.1\frac{1}{3}....1\frac{1}{2017}=\frac{3}{2}.\frac{4}{3}....\frac{2018}{2017}=\frac{2018}{2}=1009\)
=\(\frac{1}{2}x\frac{2}{3}x...x\frac{2017}{2018}\)
=\(\frac{1}{2018}\)
bạn trừ ra là đc
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{2017}\right)\cdot\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2016}{2017}\cdot\frac{2017}{2018}\)
\(=\frac{1\cdot2\cdot3\cdot....\cdot2016\cdot2017}{2\cdot3\cdot4\cdot....\cdot2017\cdot2018}\)
\(=\frac{1}{2018}\)
\(\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{9}\right)\cdot\left(1-\frac{1}{16}\right)\cdot...\cdot\left(1-\frac{1}{444}\right)\cdot\left(1-\frac{1}{169}\right)\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{143}{144}\cdot\frac{168}{169}\)
\(=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot....\cdot\frac{11.13}{12.12}\cdot\frac{12.14}{13.13}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot11\cdot12}{2\cdot3\cdot4\cdot...\cdot12\cdot13}\cdot\frac{3\cdot4\cdot5\cdot....\cdot13\cdot14}{2\cdot3\cdot4\cdot....\cdot12\cdot13}\)
\(=\frac{1}{13}\cdot\frac{14}{2}=\frac{7}{13}\)
Ta có
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times....\times\left(1-\frac{1}{10}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times....\times\frac{9}{10}\)
\(=\frac{1}{10}\)
\(=\left(\frac{1}{2}\right).\left(\frac{2}{3}\right).\left(\frac{3}{4}\right)....\left(\frac{2017}{2018}\right)\)
\(=\frac{1.2.3....2017}{2.3.4...2018}\)
\(=\frac{1}{2018}\)
\(=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot...\cdot\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2017}{2018}\)
\(=\frac{1}{2018}\)