Question

50 БАЛІВ ТЕРМІНОВО ЗНАЙДІТЬ СУМУ​

Answer 1

$\frac{1}{\sqrt{1} +\sqrt{2}}+\frac{1}{\sqrt{2} +\sqrt{3}}+\frac{1}{\sqrt{3} +\sqrt{4}}+...+\frac{1}{\sqrt{2019} +\sqrt{2020}} =\\\\=\frac{1*(\sqrt{1}-\sqrt{2})}{(\sqrt{1} +\sqrt{2})(\sqrt{1}-\sqrt{2})}+\frac{1*(\sqrt{2}-\sqrt{3)}}{(\sqrt{2} +\sqrt{3})(\sqrt{2}-\sqrt{3})}+\frac{1*(\sqrt{3}-\sqrt{4})}{(\sqrt{3} +\sqrt{4})(\sqrt{3} -\sqrt{4})}+...+\frac{1*(\sqrt{2019}-\sqrt{2020})}{(\sqrt{2019} +\sqrt{2020})(\sqrt{2019}-\sqrt{2020}}=$

= $\frac{1-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+\frac{\sqrt{3}-\sqrt{4}}{-1}+...+\frac{\sqrt{2019}-\sqrt{2020}}{-1}=\\\\=\sqrt{2}-1+\sqrt{3} -\sqrt{2}+\sqrt4} -\sqrt{3}+...+\sqrt{2020}-\sqrt{2019}}=\sqrt{2020}-1$

Answer 2

Ответ:

$\frac{ \sqrt{2020}-1 }{2019}$

Объяснение:

Домножим и числитель и знаменатель на знаменатель, только не √a+√b, а √a-√b.

$\frac{ \sqrt{1} - \sqrt{2} }{1 - 2} + \frac{ \sqrt{2} - \sqrt{3} }{2 -3} + \frac{ \sqrt{3} - \sqrt{4} }{3 - 4} ... \frac{ \sqrt{2018} - \sqrt{2019} }{2018 - 2019} + \frac{ \sqrt{2019} - \sqrt{2020} }{2019 - 2020} = \frac{1 - \sqrt{2020} }{1 - 2020} = \frac{1 - \sqrt{2020} }{-2019}$