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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 @@

\(F=-\dfrac{1}{1.2}-\dfrac{1}{2.3}-...-\dfrac{1}{2014.2015}-\dfrac{1}{2015.2016}\)

\(\Rightarrow-F=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2014.2015}+\dfrac{1}{2015.2016}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}+\dfrac{1}{2015}-\dfrac{1}{2016}=1-\dfrac{1}{2016}=\dfrac{2015}{2016}\)\(\Rightarrow F=\dfrac{-2015}{2016}\)

Giải:

\(F=\dfrac{-1}{2016.2015}-\dfrac{1}{2015.2014}-\dfrac{1}{2014.2013}-\dfrac{1}{2013.2012}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(\Leftrightarrow F=-\left(\dfrac{1}{2016.2015}+\dfrac{1}{2015.2014}+\dfrac{1}{2014.2013}+\dfrac{1}{2013.2012}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)

\(\Leftrightarrow F=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2012.2013}+\dfrac{1}{2013.2014}+\dfrac{1}{2014.2015}+\dfrac{1}{2015.2016}\right)\)

\(\Leftrightarrow F=-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)

\(\Leftrightarrow F=-\left(\dfrac{1}{1}-\dfrac{1}{2016}\right)\)

\(\Leftrightarrow F=-\dfrac{2015}{2016}\)

Vậy ...

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

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

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

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

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