Ta thấy rằng \(\dfrac{2}{3^2}\)<\(\dfrac{2}{3}\)\(\) ; \(\dfrac{2}{5^2}\)<\(\dfrac{2}{3.5}\) ; \(\dfrac{2}{7^2}\)<\(\dfrac{2}{5.7}\) ; ... ; \(\dfrac{2}{2017^2}\)<\(\dfrac{2}{2015.2017}\).
=> A<\(\dfrac{2}{3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2015.2017}\)\(\)=\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)=1-\(\dfrac{1}{2017}\)=\(\dfrac{2016}{2017}\)<\(\dfrac{2016}{4036}\)=\(\dfrac{504}{1009}\)
Vậy A<\(\dfrac{504}{1009}\)
Mình lộn chỗ so sánh rồi nên bài giải của mình sai rồi bạn ơi
Mong bạn thông cảm
