Question

B = \(\dfrac{4+\sqrt{3}}{\sqrt{1}+\sqrt{3}}+\dfrac{6+\sqrt{8}}{\sqrt{3}+\sqrt{5}}+...+\dfrac{2n+\sqrt{n^2-1}}{\sqrt{n-1}+\sqrt{n+1}}+\dfrac{240+\sqrt{14399}}{\sqrt{119}+\sqrt{121}}\)

Answer 1

ta có : \(\dfrac{2n+\sqrt{n^2-1}}{\sqrt{n-1}+\sqrt{n+1}}=\dfrac{\left(2n+\sqrt{n^2-1}\right)\left(\sqrt{n-1}+\sqrt{n+1}\right)}{-2}\)

\(=\dfrac{2n\sqrt{n-1}+2n\sqrt{n+1}+\left(n-1\right)\sqrt{n+1}+\left(n+1\right)\sqrt{n-1}}{-2}\) \(=\dfrac{\sqrt{n-1}\left(3n+1\right)+\sqrt{n+1}\left(3n-1\right)}{-2}\)

chung mẫu hết rồi cộng lại

Answer 2

lm lại nha :

ta có : \(\dfrac{2n+\sqrt{n^2-1}}{\sqrt{n-1}+\sqrt{n+1}}\) \(=\dfrac{\left(2n+\sqrt{n^2-1}\right)\left(\sqrt{n+1}-\sqrt{n-1}\right)}{2}\)

\(=\dfrac{2n\sqrt{n+1}-2n\sqrt{n-1}+\left(n+1\right)\sqrt{n-1}-\left(n-1\right)\sqrt{n+1}}{2}\)

\(=\dfrac{\left(n+1\right)\sqrt{n+1}-\left(n-1\right)\sqrt{n-1}}{2}\) cộng lại ...................

Answer 3

Mysterious Person giúp mk

Answer 4

sữa đề chút nha

ta có : \(\dfrac{2n+2\sqrt{n^2-1}}{\sqrt{n-1}+\sqrt{n+1}}=\dfrac{\left(\sqrt{n-1}+\sqrt{n+1}\right)^2}{\sqrt{n-1}+\sqrt{n+1}}=\sqrt{n-1}+\sqrt{n+1}\)

nếu không phải zầy thì mk lm lại