Răspuns:
Explicație pas cu pas:
[tex]S=...=\dfrac{\sqrt{2}-1 }{(\sqrt{2}+1)(\sqrt{2}-1)} +\dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2} )(\sqrt{3}-\sqrt{2})} +...+\dfrac{\sqrt{x+1}-\sqrt{x}}{(\sqrt{x+1}+\sqrt{x} )(\sqrt{x+1}-\sqrt{x})}=\\=\dfrac{\sqrt{2}-1 }{(\sqrt{2})^2-1^2} +\dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2} +...+\dfrac{\sqrt{x+1}-\sqrt{x}}{(\sqrt{x+1})^2-(\sqrt{x})^2}=\\[/tex][tex]=\dfrac{\sqrt{2}-1 }{2-1} +\dfrac{\sqrt{3}-\sqrt{2}}{3-2} +...+\dfrac{\sqrt{x+1}-\sqrt{x}}{x+1-x}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{x+1}-\sqrt{x}=\sqrt{x+1}-1[/tex]
Deci, S=√(x+1) - 1.
b) S=2018, ⇒ √(x+1) - 1=2018, ⇒ √(x+1) = 2018+1, ⇒√(x+1) = 2019 |², ⇒ x+1=2019², ⇒ x=2019²-1=4076361 - 1 = 4076360.
