Question

A2 va rog dacă poate cineva sa mă ajute

Answer 1

Răspuns:

3√11 - 1

Explicație pas cu pas:

$\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{98}+\sqrt{99}}=\dfrac{1}{\sqrt{2}+\sqrt{1}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{99}+\sqrt{98}}=\\=\dfrac{\sqrt{2}-\sqrt{1}}{(\sqrt{2}+\sqrt{1})(\sqrt{2}-\sqrt{1})}+\dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}+...+\dfrac{\sqrt{99}-\sqrt{98}}{(\sqrt{99}+\sqrt{98})(\sqrt{99}-\sqrt{98})}=$

$=\dfrac{\sqrt{2}-\sqrt{1}}{(\sqrt{2})^2-(\sqrt{1})^2}+\dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2}+...+\dfrac{\sqrt{99}-\sqrt{98}}{(\sqrt{99})^2-(\sqrt{98})^2}=\\=\dfrac{\sqrt{2}-\sqrt{1}}{2-1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+...+\dfrac{\sqrt{99}-\sqrt{98}}{99-98}=\\=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{99}-\sqrt{98}=\sqrt{99}-\sqrt{1}=3\sqrt{11}-1.$