Question

Exercițiul 17 punctele a și b

Answer 1

[tex]\text{Mai intai sa observam ca este un sistem omogen(daca admite solutia }\\ \text{(a,b) atunci admite si solutie (b,a))}\\ a){ \left \{ {{xy+x+y=5|\cdot 2} \atop {x^2+y^2=5}} \right.\Leftrightarrow \left \{ {{2xy+2x+2y=10} \atop {x^2+y^2=5} \right.} \\ \text{Adunand relatiile obtinem :} x^2+y^2+2xy+2x+2y=15|+1\\ x^2+y^2+2xy+2x+2y+1=16\\ (x+y+1)^2=16\\ |x+y+1|=4\\ \text{Deci avem de analizat doua cazuri:}\\ i)x+y+1=4,\text{ adica }x+y=3\text{ de unde x=3-y}\\ (3-y)^2+y^2=5\\ 9-6y+y^2+y^2=5[/tex][tex]2y^2-6y+4=0\\ y^2-3y+2=0\\ (y-1)(y-2)=0\Rightarrow y\in \{1,2\},\text{de unde obtinem solutiile :}(1,2),(2,1)\\ ii)x+y+1=-4\Rightarrow x+y=-5,\text{ deci x=-5-y}\\ (-5-y)^2+y^2=5\\ 25+10y+y^2+y^2=5\\ 2y^2+10y+20=0\\ y^2+5y+10=0\\ \Delta =25-40=-20\ \textless \ 0 \text{deci nu are solutii}\\ \text{Prin urmare }S:(x,y)\in \{(1,2),(2,1)\}[/tex][tex]b)\text{Observam ca si acesta este un sistem omogen si }x,y\neq 0\\ \dfrac{x}{y}+\dfrac{y}{x}=\dfrac{5}{2}|\cdot 2xy\\ 2x^2+2y^2=5xy\\ 2x^2-5xy+2y^2=0\\ 2x^2-4xy-xy+2y^2=0\\ 2x(x-2y)-y(x-2y)=0\\ (x-2y)(2x-y)=0\\ \text{Iar distingem doua cazuri posibile }\\ i)x-2y=0\Rightarrow x=2y\\ \text{Inlocuim in a doua ecuatie si obtinem:}\\ (2y)^2\cdot y+2y(y^2)=6\\ 4y^3+2y^3=6\\ 6y^3=6\\ y^3=1\Rightarrow y=1\text{ si }x=2 [/tex][tex]ii)2x-y=0\Rightarrow x=\dfrac{y}{2}\\ \dfrac{y^2}{4}\cdot y+y^2\cdot \dfrac{y}{2}=6|\cdot 4\\ y^3+2y^3=24\\ 3y^3=24\\ y^3=8\Rightarrow y=2\text{ si }x=1\\ \text{Prin urmare }S:(x,y)\in \{(1,2),(2,1)\}[/tex]

Answer 2

Sper sa se inteleaga pozele .Ignora ce e in a 2-a poza sus-dreapta :)