Răspuns:
Explicație pas cu pas:
[tex]Fie~\frac{1}{x^{2}}=a,~iar~\frac{1}{y^{2}} =b.~atunci~obtinem~sisyemul\\\left \{ {{5a-3b=9,~ |*2} \atop {4a+2b=16,~|*3}} \right. ~obtinem~\left \{ {{10a-6b=18} \atop {12a+6b=48}} \right.[/tex]
Adunand ecuatiile obtinem: 22a=66, ⇒a=66:22=3. Acum inlocuim in ecuatia a doua (la dorinta,,,), ⇒4·3+2b=16, ⇒2b=4, ⇒b=2.
[tex]Deci~\frac{1}{x^{2}} =3,~x^{2}=\frac{1}{3} ,~deci~x_{1}=-\sqrt{\frac{1}{3} }=-\frac{1}{\sqrt{3} } =-\frac{\sqrt{3} }{3},~iar~x_{2}= \frac{\sqrt{3} }{3}.\\Si~\frac{1}{y^{2}} =2,~y^{2}=\frac{1}{2} ,~deci~y_{1}=-\sqrt{\frac{1}{2} }=-\frac{1}{\sqrt{2} } =-\frac{\sqrt{2} }{2},~iar~y_{2}= \frac{\sqrt{2} }{2}.\\Solutiile~S=(-\frac{\sqrt{3} }{3};-\frac{\sqrt{2} }{2}); (-\frac{\sqrt{3} }{3};\frac{\sqrt{2} }{2}); (\frac{\sqrt{3} }{3};-\frac{\sqrt{2} }{2}); (\frac{\sqrt{3} }{3};\frac{\sqrt{2} }{2});[/tex]
