Question

Необходимо решить сложную систему из двух уравнений: / | x^4+x^2*y^2+y^4=133 { | x^2-xy+y^2=7 \

Answer 1

Можно так
$\left \{ {{x^4+x^2y^2+y^4=133} \atop {x^2-xy+y^2=7}} \right. \Rightarrow \left \{ {{(x^4+2x^2y^2+y^4)-x^2y^2=133} \atop {x^2+y^2-xy=7}} \right. \Rightarrow$
$\Rightarrow\left \{ {{(x^2+y^2)^2-x^2y^2=133} \atop {x^2+y^2-xy=7}} \right.$
Введем обозначения $x^2+y^2=m;xy=n.$
$\left \{ {{m^2-n^2=133} \atop m-n=7}} \right. \Rightarrow\left \{ {{(m-n)(m+n)=133} \atop m-n=7}} \right. \Rightarrow\{ {{7(m+n)=133} \atop m-n=7}} \right. \Rightarrow\{ {{m+n=19} \atop m-n=7}} \right. \Rightarrow$
$\Rightarrow\{ {{m=13} \atop n=6}} \right.$
$\left \{ {{x^2+y^2=13;} \atop {xy=6 }} \right. \Rightarrow\left \{ {{x^2+y^2=13;} \atop {2xy=12 }} \right. \Rightarrow \{ {{x^2+2xy+y^2=25;} \atop {xy=6 }} \right. \Rightarrow\{ {{(x+y)^2=25;} \atop {xy=6 }} \right. \Rightarrow$
$\left \{ {{ \left \{ {{x+y=5} \atop {xy=6}} \right. } \atop { \left \{ {{x+y=-5} \atop {xy=6}} \right. }} \right. \Rightarrow \left \{ {{ \left \{ {{x=2;y=3;} \atop {x=3;y=2;}} \right. } \atop { \left \{ {{x=-2;y=-3;} \atop {x=-3;y=-2;}} \right. }} \right.$