Aritmética

por Serg.io Qua 16 Jan 2019, 09:58

Sendo x#y = √(x^2 + 3 x y + y^2 - 2 x - 2 y + 4)/(x y + 4), ache o valor de ((...((2007#2006)#2005)#...)#1).

Gabarito:√(15)/9

por poisedom Dom 03 Fev 2019, 16:46

Suponha que ((2007#2006)#2005)#...)#3 seja k.
Assim temos
k#2= \dfrac{\sqrt{k^2+6k+4-2k-4+4}}{2k+4}= \dfrac{\sqrt{(k+2)^2}}{2(k+2)}=\dfrac{{|k+2|}}{2(k+2)}=\dfrac{1}{2}\forall k>-2

Agora basta calcular \frac{1}{2} # 1.

Que é

\frac{1}{2} # 1=\dfrac{\sqrt{\left(\dfrac{1}{2}\right)^2+3\cdot \dfrac{1}{2}\cdot 1+1^2-2\cdot \dfrac{1}{2}-2\cdot 1+4}}{\dfrac{1}{2}\cdot 1+4}.

\frac{1}{2} # 1=\dfrac{\sqrt{\dfrac{1}{4}+ \dfrac{3}{2}+1-1-2+4}}{\dfrac{1}{2}+4}

\frac{1}{2} # 1=\dfrac{\sqrt{\dfrac{1}{4}+ \dfrac{3}{2}+2}}{\dfrac{1}{2}+4}

\frac{1}{2} # 1=\dfrac{\sqrt{\dfrac{1+6+8}{4}}}{\dfrac{1+8}{2}} =\dfrac{\sqrt{\dfrac{15}{4}}}{\dfrac{9}{2}}=\dfrac{\dfrac{\sqrt{15}}{2}}{\dfrac{9}{2}}=\dfrac{\sqrt{15}}{9}

Aritmética

Aritmética

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