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\(G=\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{2}{7}\right)\left(1-\dfrac{3}{7}\right)\cdot...\cdot\left(1-1\dfrac{2}{7}\right)\left(1-1\dfrac{3}{7}\right)\\ =\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{2}{7}\right)\left(1-\dfrac{3}{7}\right)\cdot...\left(1-\dfrac{7}{7}\right)...\left(1-1\dfrac{2}{7}\right)\left(1-1\dfrac{3}{7}\right)\\ =\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{2}{7}\right)\left(1-\dfrac{3}{7}\right)\cdot...\cdot0\cdot...\cdot\left(1-1\dfrac{2}{7}\right)\left(1-1\dfrac{3}{7}\right)\\ =0\\ VậyG=0\)
\(G=\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{2}{7}\right)\left(1-\dfrac{3}{7}\right)...\left(1-1\dfrac{2}{7}\right)\left(1-1\dfrac{3}{7}\right)\)
\(\Leftrightarrow G=\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{2}{7}\right)\left(1-\dfrac{3}{7}\right)....\left(1-\dfrac{7}{7}\right)...\left(1-1\dfrac{2}{7}\right)\left(1-1\dfrac{3}{7}\right)\)
\(\Leftrightarrow G=\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{2}{7}\right)\left(1-\dfrac{3}{7}\right)...0....\left(1-1\dfrac{2}{7}\right)\left(1-1\dfrac{3}{7}\right)\)
\(\Leftrightarrow G=0\)
Vậy \(G=0\)
a) \(\left(\dfrac{2}{3}-\dfrac{1}{2}-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}-\dfrac{1}{7}\right)\)
\(=-\dfrac{1}{6}\cdot\dfrac{17}{28}\)
\(=-\dfrac{17}{168}\)
b) \(\left(\dfrac{15}{21}\div\dfrac{5}{7}\right)\div\left(\dfrac{6}{5}\div2\right)\)
\(=1\div\dfrac{3}{5}\)
\(=\dfrac{5}{3}\)
1: Ta có: \(23\dfrac{1}{4}\cdot\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
\(=\dfrac{93}{4}\cdot\dfrac{7}{5}-\dfrac{53}{4}\cdot\dfrac{7}{5}\)
\(=\dfrac{7}{5}\cdot10=14\)
2: Ta có: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)
\(=\dfrac{12+8-3}{12}\cdot\dfrac{1}{400}\)
\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)
Ta có : \(\left(1\dfrac{2}{3}\right)\left(1\dfrac{2}{5}\right).....\left(1\dfrac{2}{2011}\right)\left(1\dfrac{2}{2013}\right)\)
\(=\dfrac{5}{3}.\dfrac{7}{5}....\dfrac{2013}{2011}.\dfrac{2015}{2013}=\dfrac{2015}{3}\)
\(B=\left(1-\dfrac{1}{7}\right).\left(1-\dfrac{2}{7}\right).\left(1-\dfrac{3}{7}\right)...\left(1-\dfrac{2011}{7}\right)\)
\(B=1-\left(\dfrac{1}{7}.\dfrac{2}{7}.\dfrac{3}{7}...\dfrac{2011}{7}\right)\)
Tự tính nốt
\(=\dfrac{6}{7}\cdot\dfrac{5}{7}\cdot\dfrac{4}{7}\cdot\dfrac{3}{7}\cdot\dfrac{2}{7}\cdot\dfrac{1}{7}\cdot\dfrac{0}{7}\cdot...\cdot\dfrac{-2007}{7}\)
=0