Question

Tính :

P = \(\dfrac{1}{1+\sqrt{2}}\) + \(\dfrac{1}{\sqrt{2}+\sqrt{3}}\) + \(\dfrac{1}{\sqrt{3}+\sqrt{4}}\) + ........ + \(\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

Thanks trước nkoa mấy man

Answer 1

\(P=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

\(P=\dfrac{\sqrt{2}-1}{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{99}+\sqrt{100}\right)\left(\sqrt{100}-\sqrt{99}\right)}\)

\(P=\dfrac{\sqrt{2}-1}{2-1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+...+\dfrac{\sqrt{100}-\sqrt{99}}{100-99}\)

\(P=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\)

\(P=-1+\sqrt{100}=-1+10=9\)

Answer 2

Áp dụng:\(\dfrac{1}{\sqrt{a}+\sqrt{a+1}}=\dfrac{\sqrt{a+1}-\sqrt{a}}{\left(\sqrt{a}+\sqrt{a+1}\right)\left(\sqrt{a+1}-\sqrt{a}\right)}=\dfrac{\sqrt{a+1}-\sqrt{a}}{a+1-a}=\sqrt{a+1}-\sqrt{a}\)