Question

Exercițiul 2 ! Va rog ( măcar o formula )

Answer 1

  2.a)

[tex]\displaystyle \bf\\ 2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8+\cdots\\ \cdots+2^{2009}+2^{2010}+2^{2011}+2^{2012}\\\\ \text{Observam suma primilor 4 termeni este divizibila cu 15: }\\\\ 2^1+2^2+2^3+2^4 = 2+4+8+16 = 30 ~\vdots ~15\\\\ \text{Vom grupa sirul in grupe de cate 4 termeni.}\\ \text{Avem voie sa facem asta deoarece avem 2012 termeni, iar }2012 ~\vdots~ 4.\\\\ \Big(2^1+2^2+2^3+2^4\Big)+\Big(2^5+2^6+2^7+2^8\Big)+\cdots\\ \cdots+\Big(2^{2009}+2^{2010}+2^{2011}+2^{2012}\Big)\\\\ [/tex]

[tex]\displaystyle \bf\\ \text{Din fiecare grupa dam factor comun:}\\\\ 1\Big(2^1+2^2+2^3+2^4\Big)+2^4\Big(2^1+2^2+2^3+2^4\Big)+\cdots\\ \cdots+2^{2008}\Big(2^1+2^2+2^3+2^4\Big) =\\\\ =\Big(2^1+2^2+2^3+2^4\Big)\Big(1 + 2^4 + \cdots + 2^{2008}\Big)=\\\\ =30\Big(1 + 2^4 + \cdots + 2^{2008}\Big) ~\vdots ~15 [/tex]


2.b)

[tex]\displaystyle \bf\\ x^2 + y^2+4x-8y+20=0\\\\ \text{Descompunem pe 20 in suma de patrate perfecte.}\\ 20 = 4 + 16\\\\ x^2 + y^2+4x-8y+4+16=0\\\\ (x^2 + 4x +4) + (y^2 -8y +16)=0\\\\ (x+2)^2(y-4)^2 = 0\\\\ \Longrightarrow~~~(x+2)^2 = 0~~~sau~~~(y-4)^2 = 0\\\\ x+2 = 0 \\\\ x = -2~~radacina ~dubla\\\\ y-4 = 0\\\\ y = 4 ~~radacina ~dubla [/tex]