Question

Chứng minh: \(\dfrac{1}{\sqrt{n}} - \dfrac{1}{\sqrt{n+1}} = \dfrac{1}{(n+1).\sqrt{n} + n.\sqrt{n+1}}\)

Answer 1

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

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

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

\(=\dfrac{1}{\sqrt{n}.\left(n+1\right)+n.\sqrt{n+1}}\) = VP

=> Đpcm