Question

$\displaystyle \bold{Rezolvati~in~R^2~sistemul~de~ecuatii:} \\ \\ \bold{ \left \{ {{|x^2-5x+7|=|y|^{|y| \cdot |x^2-5x+7|}} \atop {|x^2-5x+7|^{ \frac{2}{|x^2-5x+7|}}+|y|^ \frac{1}{|y|} =2 }}} \right. }$

Answer 1

$\displaystyle Evident~y \neq 0 ~si~x^2-5x+7 \neq 0~(ceea~ce~era~oricum~adevarat)\\ \\ Din~|x^2-5x+7|=|y|^{|y||x^2-5x+7|} ~rezulta~\\ \\ | x^2-5x+7|^{ \frac{2}{|x^2-5x+7|} }=|y|^{2|y|} \\ \\A ~doua~ecuatie~devine:~|y|^{2|y|}+|y|^{ \frac{1}{|y|}}=2. \\ \\ Vom~analiza~trei~cazuri: \\ \\ i)~Daca~|y| \in (0;1) \Rightarrow |y|^{2|y|}\ \textless \ 1~si~|y|^{ \frac{1}{|y|} }\ \textless \ 1,~deci \\ \\ |y|^{2|y|}+|y|^{ \frac{1}{|y|}}<2,~contradictie! \\ \\ ii)~Daca~|y|=1,~relatia~este~satisfacuta.$

$\displaystyle iii)~Daca~|y| \in (1; \infty) \Rightarrow |y|^{2|y|}\ \textgreater \ 1~si~|y|^{ \frac{1}{|y|} }\ \textgreater \ 1,~deci \\ \\ |y|^{2|y|}+|y|^{ \frac{1}{|y|}}\ \textgreater \ 2,~contradictie! \\ \\ Prin~urmare~convine~doar~|y|=1.~Prima~ecuatie~devine: \\ \\ |x^2-5x+7|=1. \\ \\ Insa~x^2-5x+7=(x- \frac{5}{2})^2+ \frac{3}{4} \ \textgreater \ 0~(lucru~care~putea~fi \\ \\ demonstrat~si~observand~ca~ \Delta=-3\ \textless \ 0~si~a=1\ \textgreater \ 0). \\ \\ Prin~urmare~|x^2-5x+7|=x^2-5x+7.$

$Deci~ x^2-5x+7=1 \Leftrightarrow x^2-5x+6=0 \Leftrightarrow (x-2)(x-3)=0. \\ \\ x_1=2~;~x_2=3. \\ \\ y_1=-1;~y_2=1. \\ \\ Solutie: \\ \\ ~(x,y) =\{2,3\} \times \{-1;1 \} =\{(2,-1);(2,1);(3,-1);(3,1) \}.$