Question

So sánh \(\dfrac{2016\cdot2018}{1999+2016\cdot2017}\) với 1

Answer 1

Giải.

Ta có : \(\dfrac{2016.2018}{1999+2016.2017}=\dfrac{2016\left(2017+1\right)}{1999+2016.2017}\)

\(=\dfrac{2016.2017+2016}{1999+2016.2017}\)

Do \(2016>1999\)

\(\Rightarrow2016.2017+2016>1999+2016.2017\)

\(\dfrac{2016.2017+2016}{1999+2016.2017}>1\)

Vậy...

tik mik nha !!!

Answer 2

Ta có:

\(\dfrac{2016.2018}{1999+2016.2017}\)= \(\dfrac{2016\left(1+2017\right)}{1999+2016.2017}\)= \(\dfrac{2016+2016.2017}{1999+2016.2017}\)

Vì \(2016>1999\) nên \(2016+2016.2017>1999+2016.2017\)

Do đó, \(\dfrac{2016+2016.2017}{1999+2016.2017}\) > 1

Vậy \(\dfrac{2016.2018}{1999+2016.2017}\) > 1