\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\)

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

\(=1-\frac{1}{2016}\)

\(=\frac{2015}{2016}\)

Phép tính trên có thể ghi ngược lại

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2014.2015}+\frac{1}{2015.2016}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}\)

=\(1-\frac{1}{2016}\)

=\(\frac{2015}{2016}\)

A= 1/1.2 + 1/2.3 +...........+ 1/2016.2015

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

  = 1 - 1/2016

  = 2015/2016

thank nhìu nha

5/6+3/4 < 19/11                           4/9 x 6/5 = 8/15

17/12-9/8 < 1/3                            15/4:3/8 > 8

Ta có: \(\frac{5}{6}+\frac{3}{4}=\frac{10}{12}+\frac{9}{12}=\frac{19}{12}\)

Vì \(12>11\)

\(\Rightarrow\frac{19}{12}< \frac{19}{11}\)

\(\Rightarrow\frac{5}{6}+\frac{3}{4}< \frac{19}{11}\)

Ta có: \(\frac{17}{12}-\frac{9}{8}=\frac{34}{24}-\frac{27}{24}=\frac{7}{24}\)

Ta lại có: \(\frac{1}{3}=\frac{8}{24}\)

Vì \(8>7\)

\(\Rightarrow\frac{7}{24}< \frac{8}{24}\)

\(\Rightarrow\frac{17}{12}-\frac{9}{8}< \frac{1}{3}\)

Ta có: \(\frac{4}{9}\times\frac{6}{5}=\frac{24}{45}=\frac{8}{15}\)

Vì \(\frac{8}{15}=\frac{8}{15}\)

 \(\Rightarrow\frac{4}{9}\times\frac{6}{5}=\frac{8}{15}\)

Ta có: \(\frac{15}{4}:\frac{3}{8}=\frac{15}{4}\times\frac{8}{3}=10\)

Vì \(10>8\)

\(\Rightarrow\frac{15}{4}:\frac{3}{8}>8\)

HOK TOT

bai nay ban viet nguoc day so lai roi giai nhu binh thuong la duoc

\(\frac{1}{2}-\frac{1}{2016.2015}-\frac{1}{2015.2014}-...-\frac{1}{3.2}\)

\(=\frac{1}{2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\right)\)

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

\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{2016}\right)\)

\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{2016}\)

\(=\frac{1}{2016}\)

\(\frac{1}{2}-\frac{1}{2016.2015}-\frac{1}{2015.2014}-...-\frac{1}{3.2}\)

\(=\frac{1}{2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2014.2015}+\frac{1}{2015.2016}\right)\)

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

\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{2016}\right)\)

\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{2016}\)

\(=0+\frac{1}{2016}=\frac{1}{2016}\)

=\(-\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2015}+\frac{1}{2014}-...-\frac{1}{2}+1\)

=\(-\frac{1}{2016}+1=\frac{2015}{2016}\)

Ta có :\(\frac{-1}{2016.2015}-\frac{1}{2015.2014}-\frac{1}{2014.2013}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

       = \(-\left(\frac{1}{2016.2015}+\frac{1}{2015.2014}+\frac{1}{2014.2013}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

       = \(-\left(\frac{1}{2016}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2014}+\frac{1}{2014}-\frac{1}{2013}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-\frac{1}{1}\right)\)

       = \(-\left(\frac{1}{2016}-1\right)\)

       = \(-\left(-\frac{2015}{2016}\right)\)

      =  \(-\frac{2015}{2016}\)

Mk làm kĩ lắm rồi. ko tích nữa mk cũng chịu bạn luôn @@

\(A=\frac{1}{2016.2015}+\frac{1}{2015.2014}+\frac{1}{2014.2013}+...+\frac{1}{1.2}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}\)

\(=1-\frac{1}{2016}=\frac{2015}{2016}\)

Vậy \(A=\frac{2015}{2016}\).

Mình viết ngược lại cho dễ làm xD

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2014\cdot2015}+\frac{1}{2015\cdot2016}\)

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

\(A=\frac{1}{1}-\frac{1}{2016}\)

\(A=\frac{2015}{2016}\)

Sai thì bỏ quá :3

= 2015/2016 nha bạn

\(\frac{1}{2016.2015}+\frac{1}{2015.2014}+...+\frac{1}{1.2}\)

\(=\frac{1}{1.2}+...+\frac{1}{2014.2015}+\frac{1}{2015.2016}\)

\(=\frac{1}{1}-\frac{1}{2}+...+\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}\)

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

\(=\frac{1}{1}-\frac{1}{2016}=\frac{2015}{2016}\)

~ Hok tốt ~

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

=1/2016-1/2

=-1007/2016

\(\frac{1}{2016.2015}+\frac{1}{2015.2014}+...+\frac{1}{1.2}\)

\(=\frac{1}{2016}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2014}+...+\frac{1}{2}-\frac{1}{1}\)

\(=\frac{1}{2016}-1\)

\(=-\frac{2015}{2016}\)

\(\left(3^2\right)^{2010}=9^{2010}=81^{1005}\)

Vì 1 khi lũy thừa lên bao nhiêu thì số tận cùng vẫn là 1 vì 1 x 1 x 1 x 1... = ......1

Nên \(81^{1005}\)có số tận cùng là 1

Vậy \(\left(3^2\right)^{2010}\)có số tận cùng là 1