So sánh :
2015.2016-1
2015.2016
và
2016.2017-1
2016.2017
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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 \(\frac{1}{2015.2016}>\frac{1}{2016.2017}\)
=>\(-\frac{1}{2015.2016}< -\frac{1}{2016.2017}\)
=>\(1-\frac{1}{2015.2016}< 1-\frac{1}{2016.2017}\)
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
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2016\cdot2017}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}=\frac{2016}{2017}\)