Question

[Trần Phúc Khang]

Ta có

\(P=\frac{\sqrt{2}}{\sqrt{2}+\sqrt{4}}+\frac{\sqrt{2}}{\sqrt{4}+\sqrt{6}}+...+\frac{\sqrt{2}}{\sqrt{2018}+\sqrt{2020}}\)

=>\(\frac{P}{\sqrt{2}}=\frac{\sqrt{4}-\sqrt{2}}{\left(\sqrt{4}-\sqrt{2}\right)\left(\sqrt{4}+\sqrt{2}\right)}+...+\frac{\sqrt{2020}-\sqrt{2018}}{\left(\sqrt{2020}-\sqrt{2018}\right)\left(\sqrt{2020}+\sqrt{2018}\right)}\)

=>\(\frac{P}{\sqrt{2}}=\frac{\sqrt{4}-\sqrt{2}}{2}+\frac{\sqrt{6}-\sqrt{4}}{2}+...+\frac{\sqrt{2020}-\sqrt{2018}}{2}\)

=> \(\frac{P}{\sqrt{2}}=\frac{\sqrt{2020}-\sqrt{2}}{2}\)

=> \(P=\sqrt{1010}-1\)

Vậy \(P=\sqrt{1010}-1\)

Answer 1

[Thế Vĩ]

cái chỗ suy ra P e kh hiểu lắm a chỉ e chi tiết với

Answer 2

[Nguyễn Linh Chi]

@Thế Vĩ@

\(P=\sqrt{2}.\frac{\sqrt{2020}-\sqrt{2}}{2}=\sqrt{2}.\frac{\sqrt{2}\left(\sqrt{1010}-1\right)}{2}=2.\frac{\sqrt{1010}-1}{2}=\sqrt{1010}-1\)