a.\(\frac{2015.2016-1}{2015.2016}=1-\frac{1}{2015.2016}\)

\(\frac{2016.2017-1}{2016.2017}=1-\frac{1}{2016.2017}\)

vì \(\frac{1}{2015.2016}>\frac{1}{2016.2017}\)

=>\(-\frac{1}{2015.2016}< -\frac{1}{2016.2017}\)

=>\(1-\frac{1}{2015.2016}< 1-\frac{1}{2016.2017}\)

ai giúp mình với

11.2+12.3+13.4+14.5+...+12015.2016+12016.2017

=1−12+12−13+13−14+14−15+...+12015−12016+12016−12017

=1−12017=20162017

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2015.2016}=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{2016}\right)=\dfrac{1}{2}-\dfrac{1}{2016.2}< \dfrac{1}{2}\left(đpcm\right)\)

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2015.2017}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2015.2017}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{2017}\right)\\ < \dfrac{1}{2}.1=\dfrac{1}{2}\)

Sửa đề: A=2015.2017 và B=2016²

Ta có: \(A=2015.2017=\left(2016-1\right)\left(2016+1\right)=2016^2-1\)

\(B=2016^2\)

\(\Rightarrow A< B\).