Ta có : \(x=\frac{2015.2016+2}{2015.2016}=\frac{2015.2016}{2015.2016}+\frac{2}{2015.2016}=1+\frac{1}{1008.2015}\)

            \(y=\frac{2016.2017+2}{2016.2017}=\frac{2016.2017}{2016.2017}+\frac{2}{2016.2017}=1+\frac{1}{1008.2017}\)

Vì \(\frac{1}{1008.2015}>\frac{1}{1008.2017}\)

=> \(1+\frac{1}{1008.2015}>1+\frac{1}{1008.2017}\)

=> \(\frac{2015.2016+2}{2015.2016}>\frac{2016.2017+2}{2016.2017}\)

=> \(x>y\)

Ta có:

x = \(\frac{2015.2016+2}{2015.2016}=\frac{2015.2016}{2015.2016}+\frac{2}{2015.2016}=1+\frac{2}{2015.2016}=1+\frac{1}{2015.1008}\)

y = \(\frac{2016.2017+2}{2016.2017}=\frac{2016.2017}{2016.2017}+\frac{2}{2016.2017}=1+\frac{2}{2016.2017}=1+\frac{1}{1008.2017}\)

Do \(\frac{1}{2015.1008}>\frac{1}{1008.2017}\) => \(1+\frac{1}{2015.1008}>1+\frac{1}{1008.2017}\)

     => x > y

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

Vì 2016x2017-\(\frac{1}{2016x2017}\)=4066272

 2017x2018-\(\frac{1}{2017x2018}\)=4070306

Mà 4066272<4070306

Nên a<b

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2016\cdot2017}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)

\(=1-\frac{1}{2017}=\frac{2016}{2017}\)