Iata o prima solutie oferita de Integrator. La impartirile de la final mi-am cam sclintit creierasul. Am pus =…=q+r apeland mai mult la intuitie decat la calcul! O sa postez si solutia (redactata ca la a VII-a) a lui PhantomR. Prima este prin impartirea cu rest iar cea de-a doua prin divizibilitate.
![\[\begin{array}{l} {\rm{In multimea Z rezolvati ecuatia: }}3{x^2}{\rm{ }} - 2xy - y - 1 = 0\\ - - - - - - - - - - - - - - - - - - - - - \\ Solutia{\rm{ 1}}\\ \left\{ \begin{array}{l} 3{x^2} - 2xy - y - 1 = 0{\rm{ }} \Leftrightarrow 3{x^2} - 2xy = y + 1 \Leftrightarrow x(3x - 2y) = y + 1 \Rightarrow x = \frac{{y + 1}}{{3x - 2y}}\\ 3{x^2} - 2xy - y - 1 = 0{\rm{ }} \Leftrightarrow 3{x^2} - 1 = 2xy + y \Leftrightarrow 3{x^2} - 1 = y(2x + 1) \Rightarrow y = \frac{{3{x^2} - 1}}{{2x + 1}} \end{array} \right.\left| \begin{array}{l} \\ x,y\\ \end{array} \right. \in {\rm{Z}}\\ {\rm{Verificam daca }}x = 0{\rm{ sau }}y = 0,{\rm{ adica daca }}x{\rm{ si/sau }}y{\rm{ }} \in {\rm{Z sau }} \in {\rm{ Z*}}{\rm{.}}\\ {\rm{Pt }}x = 0{\rm{ }}\left\{ {y = \frac{{3{x^2} - 1}}{{2x + 1}} \Rightarrow y = \frac{{ - 1}}{1} \Rightarrow y = - 1} \right. \Rightarrow {\rm{ Daca }}x = 0{\rm{ atunci }}y \in {\rm{Z*}}\\ {\rm{Pt }}y = 0{\rm{ }}\left\{ {x = \frac{{y + 1}}{{3x - 2y}} \Rightarrow x = \frac{1}{{3x}} = \frac{1}{{3\left( {\frac{1}{3}} \right)}} = \frac{1}{{\frac{3}{3}}} = \frac{1}{1} = 1{\rm{ }} \Rightarrow {\rm{ }}x{\rm{ = }}\frac{1}{3} \Rightarrow x \in {\rm{Q si }}x \notin {\rm{Z }} \Rightarrow {\rm{y}} \ne {\rm{0}}} \right.{\rm{ (1)}}\\ {\rm{Daca }}y \in {\rm{Z* atunci restul impartirii }}\frac{{3{x^2} - 1}}{{2x + 1}}{\rm{ este 0}}{\rm{.}}\\ x{\rm{ in relatia }}y = \frac{{3{x^2} - 1}}{{2x + 1}}{\rm{ poate fi }}\left\{ \begin{array}{l} x = 0\\ x = 2k{\rm{ }}(par)\\ x = 2k + 1{\rm{ }}(impar) \end{array} \right.\\ {\rm{Pt }}x = 2k{\rm{ }}\left\{ {y = \frac{{3{x^2} - 1}}{{2x + 1}} \Rightarrow y = \frac{{3{{(2k)}^2} - 1}}{{2(2k) + 1}} = \frac{{12{k^2} - 1}}{{4k + 1}} = ... = q + r{\rm{ }} \Rightarrow {\rm{ doar pt }}k = 0{\rm{ putem avea }}r = 0;} \right.{\rm{ pt }}k = 0{\rm{ }} \Rightarrow 2k = x = 0{\rm{ (2)}}\\ {\rm{Pt }}x = 2k + 1{\rm{ }}\left\{ {y = \frac{{3{x^2} - 1}}{{2x + 1}} \Rightarrow y = \frac{{3{{(2k + 1)}^2} - 1}}{{2(2k + 1) + 1}} = \frac{{3(4{k^2} + 1) - 1}}{{2(2k + 1) + 1}} = \frac{{12{k^2} + 4}}{{4k + 3}} = ... = q + r{\rm{ }} \Rightarrow {\rm{ doar pt }}k = - 1{\rm{ putem avea }}r = 0;} \right.{\rm{ pt }}k = - 1{\rm{ }} \Rightarrow x = 2k + 1 = - 1{\rm{ si }}y = \frac{{3{x^2} - 1}}{{2x + 1}}{\rm{ = - 2 (3)}}\\ \\ {\rm{ Din (1)}}{\rm{, (2) si (3) }} \Rightarrow x,y \in \left\{ {(0, - 1),( - 1, - 2)} \right\} \end{array}\]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-e8379f718bc7dfa5fdf9aef01e974f47_l3.png "Rendered by QuickLaTeX.com")
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Solutia{\rm{ 2}}\\
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\left\{ {3{x^2} - 2xy - y - 1 = 0{\rm{ }} \Leftrightarrow 3{x^2} - 1 = 2xy + y \Leftrightarrow 3{x^2} - 1 = y(2x + 1) \Rightarrow y = \frac{{3{x^2} - 1}}{{2x + 1}}} \right.\\
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{\rm{Daca }}y \in {\rm{Z* atunci (}}3{x^2} - 1) \vdots (2x + 1){\rm{ cu conditia ca numitorul sa nu fie nul}}{\rm{. }}\\
{\rm{Verificam daca }}2x + 1 \ne 0;{\rm{ }}y = \frac{{3{x^2} - 1}}{{2x + 1}} = \frac{{3{{\left( { - \frac{1}{2}} \right)}^2} - 1}}{{2\left( { - \frac{1}{2}} \right) + 1}} = \frac{{3\left( {\frac{1}{4}} \right) - 1}}{{ - \frac{2}{2} + 1}} = \frac{{\frac{3}{4} - \frac{4}{4}}}{{ - 1 + 1}} = \frac{{ - \frac{1}{4}}}{0} = 0;{\rm{ }} \Rightarrow {\rm{Doar pt }}x = \left( { - \frac{1}{2}} \right){\rm{, }}y = 0;{\rm{ Dar }}x \notin {\rm{Q deci }}2x + 1 \ne 0\\
{\rm{Aplicam divizibilitatea]
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Da pe 2xy si pe y dincolo si da pe y factor comun si scoate pe y in functie de x si iti da o fractie care apartine lui Z. Apoi foloseste divizibilitatea ca la clasa a5a si iti da .
*** QuickLaTeX cannot compile formula: \[\begin{array}{l} {\rm{In multimea Z rezolvati ecuatia] *** Error message: \begin{array} on input line 9 ended by \end{document}. leading text: \end{document} Improper \prevdepth. leading text: \end{document} Missing $ inserted. leading text: \end{document} Missing } inserted. leading text: \end{document} Missing } inserted. leading text: \end{document} Missing } inserted. leading text: \end{document} Missing \cr inserted. leading text: \end{document} Missing $ inserted. leading text: \end{document} You can't use `\end' in internal vertical mode. leading text: \end{document} \begin{array} on input line 9 ended by \end{document}. leading text: \end{document} Missing } inserted. leading text: \end{document} Emergency stop.Dupa cum ti’a spus si PhantomR. Fractia este intreaga si folosesti divizibilitatea.