Đặt: \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{7}{8}.....\frac{2013}{2014}\) (1)

Ta thấy \(A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.\frac{8}{9}.....\frac{2014}{2015}\)

Do đó nhân vế với vế, ta được: 

\(A^2< \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}.\frac{7}{8}.\frac{8}{9}.....\frac{2013}{2014}.\frac{2014}{2015}\)

\(\Rightarrow A^2< \frac{1}{2015}\)

Mặt khác, \(A>\frac{1}{2}.\frac{4}{5}.\frac{6}{7}.\frac{8}{9}.....\frac{2014}{2015}\) (2)

Từ (1) và (2), ta được: 

\(A^2>\frac{1}{4}.\left(\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}.\frac{7}{8}.\frac{8}{9}.....\frac{2013}{2014}.\frac{2014}{2015}\right)\)

\(\Rightarrow A^2>\frac{1}{4}.\frac{3}{2015}\Rightarrow A^2>\frac{3}{8060}>\frac{1}{4028}\)

Xét Tử số của A ta có:

\(2014+\frac{2013}{2}+\frac{2012}{3}+....+\frac{2}{2013}=1+\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+....+\left(\frac{1}{2014}+1\right)\)\(TS=\frac{2015}{2}+\frac{2015}{3}+....+\frac{2015}{2014}+\frac{2015}{2015}\)

\(TS=2015.\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}\right)\)

\(A=\frac{2015.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)}{\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}\right)}=2015\)

toán lớp 8 dễ quá vậy

A=2015

hình như thế

xét mẫu(chỗ 1/2014 sửa lại thành 2/2014)

=(1/2015+1)+(2/2014+1)+...+(2013/3+1)+(2014/2+1)+(2015/1-2014)

=2016/2015+2016/2014+...+2016/3+2016/2+1

=2016.(1/2016+1/2015+...+1/4+1/3+1/2)

=> A= 1/2016

mún dễ hỉu hơn hãy gửi tin nhắn cho mik

1 phan 2016. cac lam de lam

A=2015

Cần cách làm thì tích nha

có 2014/1+2013/2+2012/3+...+2/2013+1/2014=[1+(2013/2)]+[1+(2012/3)]+...+[1+(2/2013)]+[1+(1/2014)]+1

=2015/2+2015/3+...+2015/2014+2015/2015=2015.[1/2+1/3+..+1/2015)

vậy (1/2+1/3+...+1/2015).x=(1/2+1/3+...+1/2015).2015

x=2015