Question

До іть будь ласочкааа
система
х⁴-у⁴=3,
х²+у²=3;

Answer 1

ответ: приложен

Объяснение:

Answer 2

$\displaystyle\bf}\star a^2-b^2=(a+b)(a-b)\star\\\\ \left \{ {{x^4-y^4=3} \atop {x^2+y^2=3}} \right. =\div\left \{ {{(x^2+y^2)(x^2-y^2)=3} \atop {x^2+y^2=3}} \right. \\\\x^2-y^2=1= \\\\+\left \{ {{x^2-y^2=1} \atop {x^2+y^2=3}} \right. = x^2+x^2+y^2-y^2=4\\\\2x^2=4= x_{1;2}=\pm\sqrt{2} =1) x=\sqrt{2} \quad; y_{1,2}=\pm1 : \quad 2) x=-\sqrt{2} ; \quad y_{3,4}=\pm1\\\\Otvet: (\sqrt{2} ;1) \quad; (\sqrt{2} ;-1) \quad; (-\sqrt{2} ;1) \quad; (-\sqrt{2} ;-1)$