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\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{9}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2}{3}.....\frac{8}{9}\)
\(\Rightarrow A=\frac{1.2.3.....8}{2.3.4.....9}=\frac{1}{9}\)
1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2
= 1/90 - ( 1/72 + 1/56 + 1/42 + 1/30 + 1/20 + 1/12 + 1/6 + 1/2)
= 1/90 - ( 1/2 + 1/6 + 1/12 + ...+ 1/72)
= 1/90 - ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9)
= 1/90 - ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9)
= 1/90 - ( 1 - 1/9)
= 1/90 - 8/9
= 1/90 - 80/90
= -79/90
1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2
= 1/90 - ( 1/72 + 1/56 + 1/42 + 1/30 + 1/20 + 1/12 + 1/6 + 1/2)
= 1/90 - ( 1/2 + 1/6 + 1/12 + ...+ 1/72)
= 1/90 - ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9)
= 1/90 - ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9)
= 1/90 - ( 1 - 1/9)
= 1/90 - 8/9
= 1/90 - 80/90
= -79/90
mk nha cac ban
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{5}-1\right)\left(\frac{1}{6}-1\right)...\left(\frac{1}{2018}-1\right)\)
\(-A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{6}\right)...\left(1-\frac{1}{2018}\right)\)
\(-A=\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot...\cdot\frac{2017}{2018}\)
\(-A=\frac{3}{2018}\)
\(A=-\frac{3}{2018}\)
\(\left(\frac{1}{4}-1\right).\left(\frac{1}{5}-1\right)...\left(\frac{1}{2017}-1\right)\left(\frac{1}{2018}-1\right)\)
= \(-\frac{3}{4}.\frac{-4}{5}....\frac{-2016}{2017}.\frac{-2017}{2018}\)
= \(\frac{\left(-3\right).\left(-4\right)....\left(-2016\right).\left(-2017\right)}{4.5...2017.2018}\)
= \(\frac{\left(-3\right).4.5.6...2016.2017}{4.5..2017.2018}\)
= \(\frac{-3}{2018}\)
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