Question

Tính:

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

Answer 1

Ta chứng minh đẳng thức \(\dfrac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}\Leftrightarrow1=\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)=k+1-k=1\) (luôn đúng)

Khi đó, ta được \(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{n-1}+\sqrt{n}}=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{n}-\sqrt{n-1}=\sqrt{n}\)