Question

[Phạm Thị Duyên]

So sánh \(A=\sqrt{2017^2-1}-\sqrt{2016^2-1}\)

            \(B=\frac{2\cdot2016}{\sqrt{2017^2-1}-\sqrt{2016^2-1}}\)

Answer 1

[Nguyễn Ngọc Khanh (Team ASL)]

\(A=\sqrt{\left(2017-1\right)\left(2017+1\right)}-\sqrt{\left(2016-1\right)\left(2016+1\right)}\)

\(=\sqrt{2016.2018}-\sqrt{2015.2017}< \sqrt{2018.2018}-\sqrt{2015.2015}\)

\(=2018-2015=3\)

\(\Rightarrow\frac{1}{A}>\frac{1}{3}\)

\(B=\frac{2.2016}{A}>\frac{2.2016}{3}=1344>3>A\)

Vậy ta được B lớn hơn A rất nhiều :))