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\(x\)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, \(1\frac{1}{2}.1\frac{1}{3}....1\frac{1}{2017}=\frac{3}{2}.\frac{4}{3}....\frac{2018}{2017}=\frac{2018}{2}=1009\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\Rightarrow\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(\Rightarrow4x+\frac{15}{16}=1\)
\(\Rightarrow4x=\frac{1}{16}\)
\(\Rightarrow x=\frac{1}{64}\)
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{10}\right)=\frac{x}{2010}\)
\(\Leftrightarrow\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{9}{10}=\frac{x}{2010}\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot....\cdot9}{2\cdot3\cdot4\cdot....\cdot10}=\frac{x}{2010}\)
\(\Leftrightarrow\frac{1}{10}=\frac{x}{2010}\)
\(\Leftrightarrow x=\frac{2010}{10}=201\)
Ta có : \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{10}\right)=\frac{x}{2010}\)
=> \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{9}{10}=\frac{x}{2010}\)
\(\Rightarrow\frac{1.2.3......9}{2.3.4.....10}=\frac{x}{2010}\)
\(\Rightarrow\frac{1}{10}=\frac{x}{2010}\)
\(\Rightarrow x=\frac{2010}{10}=201\)
\(=\frac{1}{10}\)
\(\left(1+\frac{1}{2}\right)\times\left(1+\frac{1}{3}\right)\times\left(1+\frac{1}{4}\right)\times....\times\left(1+\frac{1}{98}\right)\times\left(1+\frac{1}{99}\right)\)
\(=\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times....\times\frac{99}{98}\times\frac{100}{99}\)
\(=\frac{3\times4\times5\times...\times99\times100}{2\times3\times4\times....\times98\times99}\)
\(=\frac{100}{2}=50\)
(1+1/2).(1+1/3).(1+1/4)....(1+1/98).(1+1/99)
=3/2.4/3.5/4...99/98.100/99
=3.4.5....99.100/2.3.4....98.99
=100/2
=50
\(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)x\left(1-\frac{1}{6}\right)\)
= \(\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}x\frac{5}{6}\)
=\(\frac{1x2x3x4x5}{2x3x4x5x6}\)
Loại 2x3x4x5 vì cả 2 vế cùng có
=\(\frac{1}{6}\)
\(\left(1-\frac{1}{2}\right)\) x \(\left(1-\frac{1}{3}\right)\)x \(\left(1-\frac{1}{4}\right)\)x \(\left(1-\frac{1}{5}\right)\)x \(\left(1-\frac{1}{6}\right)\)
\(=\)\(\frac{1}{2}\) x \(\frac{2}{3}\)x \(\frac{3}{4}\)x \(\frac{4}{5}\)x \(\frac{5}{6}\)
\(=\)\(\frac{1}{6}\)
=\(\frac{1}{2}x\frac{2}{3}x...x\frac{2017}{2018}\)
=\(\frac{1}{2018}\)
bạn trừ ra là đc
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{2017}\right)\cdot\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2016}{2017}\cdot\frac{2017}{2018}\)
\(=\frac{1\cdot2\cdot3\cdot....\cdot2016\cdot2017}{2\cdot3\cdot4\cdot....\cdot2017\cdot2018}\)
\(=\frac{1}{2018}\)
\(A=\frac{2015+2016+2017}{2014+2015+2016+2017+2018}x1000\)