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\(b.\)ghi lại đề nha bn
\(=\frac{2.2306}{1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{230.231}{2}}}\)
\(=\frac{2.2306}{1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{230.231}}\)
\(=\frac{2.2306}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{230.231}\right)}\)
\(=\frac{2.2306}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{230}-\frac{1}{231}\right)}\)
\(=\frac{2.2306}{1+2.\left(\frac{1}{2}-\frac{1}{231}\right)}\)
\(=\frac{2.2306}{1+1-\frac{2}{231}}\)
\(=\frac{2.2306}{2-\frac{2}{231}}\)
\(=\frac{2.2306}{2\left(1-\frac{1}{231}\right)}\)
\(=\frac{2306}{1-\frac{1}{231}}\)
mình nha bn thanks nhìu <3
a) \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+...+\frac{1}{2016}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\left(\frac{2015}{2}+1\right)+...+\left(\frac{1}{2016}+1\right)+1}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2017}{2}+...+\frac{2017}{2016}+\frac{2017}{2017}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{2017.\left(\frac{1}{2}+...+\frac{1}{2016}+\frac{1}{2017}\right)}\)
\(=\frac{1}{2017}\)
\(B=\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}\)
\(B=2016+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}\)
\(B=1+\left(\frac{2015}{2}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)\)
\(B=\frac{2017}{2017}+\frac{2017}{2}+...+\frac{2017}{2015}+\frac{2017}{2016}\)
\(B=2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)\)
\(\frac{B}{A}=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{2}{2017}}=2017\)
\(B=\frac{1}{1.2}=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(B=\left(1-\frac{1}{2018}\right)-\left(\frac{1}{2}-\frac{1}{2}\right)-...-\left(\frac{1}{2017}-\frac{1}{2017}\right)\)
\(B=1-\frac{1}{2018}=\frac{2017}{2018}\)
Vậy \(B=\frac{2017}{2018}\)
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\left(\frac{2009}{2}+1\right)+\left(\frac{2008}{3}+1\right)+....+\left(\frac{1}{2010}+1\right)}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2011}{2}+\frac{2011}{3}+....+\frac{2011}{2010}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{2011\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\right)}\)
\(=\frac{1}{2011}\)