Question

\(B=\dfrac{1}{\sqrt{x}+\sqrt{x+1}}+\dfrac{1}{\sqrt{x+1}+\sqrt{x+2}}+...+\dfrac{1}{\sqrt{x+2015}+\sqrt{x+2016}}\)với x = 2017

Answer 1

B = \(\dfrac{1}{\sqrt{x}+\sqrt{x+1}}+\dfrac{1}{\sqrt{x+1}+\sqrt{x+2}}+...+\dfrac{1}{\sqrt{x+2015}+\sqrt{x+2016}}\)

B = \(\dfrac{\sqrt{x}-\sqrt{x+1}}{x-x-1}+\dfrac{\sqrt{x+1}-\sqrt{x+2}}{x+1-x-2}+...+\dfrac{\sqrt{x+2015}-\sqrt{x+2016}}{x+2015-x-2016}\)

B = \(\dfrac{\sqrt{x}-\sqrt{x+1}}{-1}+\dfrac{\sqrt{x+1}-\sqrt{x+2}}{-1}+...+\dfrac{\sqrt{x+2015}-\sqrt{x+2016}}{-1}\)

B = \(-\sqrt{x}+\sqrt{x+1}-\sqrt{x+1}+\sqrt{x+2}-...-\sqrt{2015}+\sqrt{2016}\)

B = \(-\sqrt{x}+\sqrt{2016}\)

Khi x = 2017

B = \(-\sqrt{2017}+\sqrt{2016}=\sqrt{2016}-\sqrt{2017}\)

Answer 2

Gợi ý: sử dụng trục căn thức.