Matematică, întrebare adresată de utilizator14, 9 ani în urmă

Vă rog să mă ajutați !

Anexe:

Răspunsuri la întrebare

Răspuns de Rayzen
5
1\ \textless \ \dfrac{1}{\sqrt3+\sqrt2}+\dfrac{1}{\sqrt4+\sqrt3}+...+\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\ \textless \ 3 \quad ($rationalizam) \\ \\ \\$1\ \textless \ \dfrac{(\sqrt3-\sqrt2)}{(\sqrt3+\sqrt2)(\sqrt3+\sqrt2)}+\dfrac{(\sqrt4-\sqrt3)}{(\sqrt4+\sqrt3)(\sqrt4-\sqrt3)}+\\ +...+\dfrac{\sqrt{n+1}-\sqrt{n}}{(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}+\sqrt{n})}\ \textless \ 3 \\ \\ \\1\ \textless \ \dfrac{\sqrt3-\sqrt2}{\sqrt3^2-\sqrt2^2}+\dfrac{\sqrt4-\sqrt3}{\sqrt4^2-\sqrt3^2} +...+\dfrac{\sqrt{n+1}-\sqrt n}{\sqrt{n+1}^2-\sqrt{n}}\ \textless \ 3

1\ \textless \ \dfrac{\sqrt3-\sqrt2}{3-2}+\dfrac{\sqrt4-\sqrt3}{4-3}+...+\dfrac{\sqrt{n+1}-\sqrt n}{n+1-n}\ \textless \ 3 \\ \\ \\ 1\ \textless \ \sqrt3-\sqrt2+\sqrt4-\sqrt3+...+\sqrt{n+1}-\sqrt n \ \textless \ 3 \\ \\ \\ 1\ \textless \ \sqrt3-\sqrt2+\sqrt4-\sqrt3+\sqrt5-\sqrt4+...+\\ \\ +\sqrt {n}-\sqrt{n-1}+\sqrt{n+1} -\sqrt n \ \textless \  3 \\ \\ \\ 1\ \textless \  -\sqrt2+\sqrt{n+1}\ \textless \ 3 \Big|+\sqrt2  \\ \\ 1+\sqrt2\ \textless \ \sqrt{n+1}\ \textless \ 3+\sqrt2\Big|^2 \\ \\ 1+2\sqrt2+2 \ \textless \  n+1\ \textless \  9+6\sqrt2+2 \Big|-1 \\ \\ 2\sqrt2+2\ \textless \  n\ \textless \  10+6\sqrt2,\quad (\sqrt2 \approx 1,41) \\ \\ 2\cdot 1,41+2\ \textless \  n\ \textless \  10+6\cdot 1,41

2,82+2\ \textless \ n\ \textless \ 10+8,46 \\ \\ 4,82\ \textless \ n\ \textless \ 18,46 \\ \\ \Rightarrow \boxed{n \in \Big\{5,6,7,8,9,10,11,12,13,14,15,16,17,18\Big\}}

Rayzen: S-a putea sa fi omis ceva, mai verific o data.
Rayzen: Am gasit, modific imediat.
Rayzen: Gata! Am corectat, acum este corect.
utilizator14: Multumesc!
Rayzen: Cu placere!
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