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\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\Rightarrow\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(\Rightarrow4x+\frac{15}{16}=1\)
\(\Rightarrow4x=\frac{1}{16}\)
\(\Rightarrow x=\frac{1}{64}\)
\(\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{9}\right)\cdot\left(1-\frac{1}{16}\right)\cdot...\cdot\left(1-\frac{1}{444}\right)\cdot\left(1-\frac{1}{169}\right)\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{143}{144}\cdot\frac{168}{169}\)
\(=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot....\cdot\frac{11.13}{12.12}\cdot\frac{12.14}{13.13}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot11\cdot12}{2\cdot3\cdot4\cdot...\cdot12\cdot13}\cdot\frac{3\cdot4\cdot5\cdot....\cdot13\cdot14}{2\cdot3\cdot4\cdot....\cdot12\cdot13}\)
\(=\frac{1}{13}\cdot\frac{14}{2}=\frac{7}{13}\)
\(A=\frac{2015+2016+2017}{2014+2015+2016+2017+2018}x1000\)
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{10}\right)=\frac{x}{2010}\)
\(\Leftrightarrow\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{9}{10}=\frac{x}{2010}\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot....\cdot9}{2\cdot3\cdot4\cdot....\cdot10}=\frac{x}{2010}\)
\(\Leftrightarrow\frac{1}{10}=\frac{x}{2010}\)
\(\Leftrightarrow x=\frac{2010}{10}=201\)
Ta có : \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{10}\right)=\frac{x}{2010}\)
=> \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{9}{10}=\frac{x}{2010}\)
\(\Rightarrow\frac{1.2.3......9}{2.3.4.....10}=\frac{x}{2010}\)
\(\Rightarrow\frac{1}{10}=\frac{x}{2010}\)
\(\Rightarrow x=\frac{2010}{10}=201\)
\(=\frac{1}{10}\)
