Question

Cine ma ajuta la cele doua?

multumesc anticipat

Answer 1

$29.~Folosim~inegalitatea~lui~Minkowski: \\ \\ \sqrt{a^2+b^2}+ \sqrt{m^2+n^2} \ge \sqrt{(a+m)^2+(b+n)^2},~cu~egalitate \\ \\ daca~an=bm~si~am+bn \ge 0. \\ \\ Luam~a=x-4,~b=y,~m=-x,~n=-y+2 \sqrt{5},~si~obtinem: \\ \\ E \ge \sqrt{(-4)^2+(2 \sqrt{5})^2}=6. \\ \\ Verificand~conditiile~cazului~de~egalitate,~observam~ca~egalitatea \\ \\ are~loc~(de~exemplu)~pentru~x=4~si~y=0. \\ \\ Deci~minimul~expresiei~este~6.$

$\displaystyle 30. ~Utilizand~binomul~lui~Newton~constatam~ca \\ \\ (1+ \sqrt{2})^n=x_n+y_n \sqrt{2}~si~(1- \sqrt{2})^n=x_n-y_n \sqrt{2}. \\ \\ Din~aceste~relatii~rezulta ~x_n= \frac{(1+ \sqrt{2})^n+(1- \sqrt{2})^n}{2}~si \\ \\ y_n= \frac{(1+\sqrt2)^n-(1- \sqrt{2})^n}{2 \sqrt{2}}. \\ \\ Deci~ \frac{x_n}{y_n}= \frac{(1+ \sqrt2)^n+(1- \sqrt2)^n}{(1+ \sqrt2)^n-(1- \sqrt2)^n} \cdot \sqrt{2}.$
$\displaystyle \frac{x_n}{y_n}= \frac{1+ \left( \frac{1- \sqrt{2} }{1+ \sqrt{2}} \right)^n}{1- \left( \frac{1- \sqrt{2}}{1+ \sqrt{2}} \right)^n}. \\ \\ Avem~\frac{1- \sqrt{2} }{1+ \sqrt{2}} \in (-1,0).~Deci~ \lim_{n \to \infty}} \left( \frac{1- \sqrt{2} }{1+ \sqrt{2}} \right)^n=0.~(Aici~fac \\ \\ observatia~ca~acest~ultim~sir~are~doua~subsiruri~dupa~cum~n~este~ \\ \\ par,~respectiv~impar.) \\ \\ Deci \lim_{n \to \infty}} \frac{x_n}{y_n}= \frac{1+0}{1-0} \cdot \sqrt{2}= \sqrt{2}.$