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8 tháng 4 2017

TA có :\(\frac{2015.2016-1}{2015.2016}=\frac{2015.2016}{2015.2016}-\frac{1}{2015.2016}=1-\frac{1}{2015.2016}\)

Ta có:\(\frac{2016.2017-1}{2016.2017}=\frac{2016.2017}{2016.2017}-\frac{1}{2016.2017}=1-\frac{1}{2016.2017}\)

Vì \(2015.2016< 2016.2017\)

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

\(\Rightarrow1-\frac{1}{2015.2016}< 1-\frac{1}{2016.2017}\)

\(\Rightarrow\frac{2015.2016-1}{2015.2016}< \frac{2016.2017-1}{2016.2017}\)

Vậy \(\frac{2015.2016-1}{2015.2016}< \frac{2016.2017-1}{2016.2017}\)

Ta có \(B=\frac{2015+2016+2017}{2016+2017+2018}\)

\(\Leftrightarrow B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)


\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018};\frac{2016}{2017}>\frac{2016}{2016+2017+2018};\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\) nên \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)

Hay \(A>B\)