Question

rút gọn biểu thức

\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+........\dfrac{1}{\sqrt{n-1}+\sqrt{n}}\)

Answer 1

\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{n-1}+\sqrt{n}}\)

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

\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{n}-\sqrt{n-1}\)

\(=\sqrt{n}-1\)

Answer 2

Ta có công thức tổng quát

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