Giải thích các bước giải:
Sửa đề:
Tính tổng $T=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{2020}+\sqrt{2019}}$
Ta có:
$T=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{2020}+\sqrt{2019}}$
$\to T=\dfrac{1}{\sqrt{2020}+\sqrt{2019}}+...+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+1}$
$\to T=\dfrac{\sqrt{2020}-\sqrt{2019}}{(\sqrt{2020}+\sqrt{2019})(\sqrt{2020}-\sqrt{2019})}+...+\dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}+\dfrac{\sqrt{2}-1}{(\sqrt{2}-1)(\sqrt{2}+1)}$
$\to T=\dfrac{\sqrt{2020}-\sqrt{2019}}{2020-2019}+...+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+\dfrac{\sqrt{2}-1}{2-1}$
$\to T=\sqrt{2020}-\sqrt{2019}+...+\sqrt{3}-\sqrt{2}+\sqrt{2}-1$
$\to T=\sqrt{2020}-1$
