Question

Va rog, sunt curios de rezolvare!

Answer 1

Se amplifica fiecare fractie cu diferenta termenilor de la numarator.

$\frac{ \sqrt{3} - \sqrt{1} }{3 - 1} + \frac{ \sqrt{5} - \sqrt{3} }{5 - 3} + \frac{ \sqrt{7} - \sqrt{5} }{7 - 5} + ...... + \frac{ \sqrt{2025} - \sqrt{2023} }{2025 - 2023} = \frac{ \sqrt{3} - \sqrt{1} }{2} + ..... + \frac{ \sqrt{2025} - \sqrt{2023} }{2} = \frac{1}{2} \times ( \sqrt{3} - \sqrt{1} + \sqrt{5} - \sqrt{3} + \sqrt{7} - \sqrt{5} + .... + \sqrt{2025} - \sqrt{2023} ) = \frac{1}{2} \times ( \sqrt{2025} - 1)$

Answer 2

$\it \dfrac{1}{1+\sqrt3} +\dfrac{1}{\sqrt3+\sqrt5} +\dfrac{1}{\sqrt5+\sqrt7}+\ ...\ +\dfrac{1}{\sqrt{2023}+\sqrt{2025}} =\\ \\ \\ =\dfrac{^{\sqrt3-1)}1}{\ \ \sqrt3+1} +\dfrac{^{\sqrt5-\sqrt3)}1}{\ \ \sqrt5+\sqrt3} +\dfrac{^{\sqrt7-\sqrt5)}1}{\ \ \sqrt7+\sqrt5}+\ ...\ +\dfrac{^{\sqrt{2025}-\sqrt{2023})}1}{\ \ \ \sqrt{2025}+\sqrt{2023}} =\\ \\ \\ =\dfrac{\sqrt3-1+\sqrt5-\sqrt3+\sqrt7-\sqrt5 +\ ...\ +\sqrt{2025}-\sqrt{2023}}{2}=\dfrac{\sqrt{2025}-1}{2}=\\ \\ \\ =\dfrac{45-1}{2}=\dfrac{44}{2}=22$