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8 tháng 6 2017

Sửa đề: Cho \(V=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\)và \(Y=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\). Tính \(\frac{V}{Y}\)

\(V=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2016}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2016}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)

\(=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)

=> \(\frac{V}{Y}=\frac{\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}}{\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}}=1\)

8 tháng 6 2017

V = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}\)

V = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)

V = \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)

V = \(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2015}+\frac{1}{2016}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)

V = \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)

V = \(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)

Vậy V : Y = \(\frac{\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}}{\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2016}}\)

( Mình nghĩ Y = 1/1009 + 1/1010 + ... + 1/2016 / Nếu Y như mình nói thì V : Y = 1 )