Question

Cel putin 2 va rog frumos.

Answer 1

Salut,

AL 9:

$a_{n+1}-a_n=\dfrac{n+2-1}{(n+2)!}=\dfrac{n+2}{(n+2)!}-\dfrac{1}{(n+2)!}=\\\\=\dfrac{n+2}{(n+2)\cdot(n+1)!}-\dfrac{1}{(n+2)!}=\dfrac{1}{(n+1)!}-\dfrac{1}{(n+2)!};\\\\a_1-a_0=\dfrac{1}{1!}-\dfrac{1}{2!};\\\\a_2-a_1=\dfrac{1}{2!}-\dfrac{1}{3!};\\\ldots\\a_{n}-a_{n-1}=\dfrac{1}{n!}-\dfrac{1}{(n+1)!}\\\\a_{n+1}-a_n=\dfrac{1}{(n+1)!}-\dfrac{1}{(n+2)!}.\\\\Adun\breve{a}m\ toate\ aceste\ rela\c{t}ii,\ membru\ cu\ membru:\\a_{n+1}-a_0=1-\dfrac{1}{(n+2)!},\ a_0=0,\ din\ enun\c{t},\ deci:\\\\a_{n+1}=1-\dfrac{1}{(n+2)!}.\ Pentru\ n=2012,\ ob\c{t}inem:\\\\a_{2013}=1-\dfrac{1}{2014!},\ r\breve{a}spunsul\ corect\ este\ \boxed{e}.$

AL 10:

$\\\\x=\dfrac{1}{2\sqrt5-4}+\sqrt2=\dfrac{1}{\sqrt{20}-\sqrt{16}}+\sqrt2=\dfrac{\sqrt{20}+\sqrt{16}}{(\sqrt{20}+\sqrt{16})(\sqrt{20}-\sqrt{16})}+\sqrt2=\\\\\\=\dfrac{2\sqrt5+4}{20-16}+\sqrt2=\dfrac{2\sqrt5+4}{4}+\sqrt2=\dfrac{\sqrt5+2}2+\sqrt2=\dfrac{\sqrt5}2+1+\sqrt2.\\\\\sqrt5>\sqrt4\Rightarrow\sqrt5>2\Rightarrow\dfrac{\sqrt5}2>1,\ deci\ \left[\dfrac{\sqrt5}2\right]=1;\\\\\[x\]=\left[\dfrac{\sqrt5}2\right]+[1]+[\sqrt2]=1+1+1=3,\ deci\ [x]=3.$

Green eyes.

Answer 2

11)

 [tex]\it \left[\dfrac{x-2}{3}\right]=\dfrac{2x-4}{5},\ \ x\geq2 \\\;\\ \\\;\\ \underline{\mathcal{R}:}[/tex]


[tex]\it Fie\ k\in\mathbb{Z}. \\\;\\ \dfrac{2x-4}{5}=k \Rightarrow x=\dfrac{5k+4}{2} \Rightarrow x=\dfrac{5k}{2} +2\ \ \ (1) \\\;\\ \\\;\\ x\geq2 \Rightarrow \dfrac{5k}{2} +2 \geq2 \Rightarrow \dfrac{5k}{2} \geq0 \Rightarrow k\geq0 \ \ \ (2)[/tex]


Ecuația devine:


[tex]\it \left[\dfrac{\dfrac{5k}{2}+2-2}{3}\right]= k \Rightarrow \left[\dfrac{5k}{6}\right]= k \Rightarrow k\leq\dfrac{5k}{6}\ \textless \ k+1 \Rightarrow \\\;\\ \\\;\\ \Rightarrow 6k\leq5k\ \textless \ 6k+6 [/tex]

Din dubla inecuație, ținând seama de relația (2), rezultă k = 0, pe care îl

substituim în relația (1) și  obținem x = 2.

Deci, S = {2}.