Salut,
AL 9:
[tex]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}.[/tex]
AL 10:
[tex]\\\\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.[/tex]
Green eyes.
