Question

Решите пожалуйста срочно!!!!!​

Answer 1

$\left \{ {{x^2+y^2+x+y=18} \atop {x^2+y^2+xy=12}} \right.\; \; \left \{ {{x^2+y^2=-x-y+18} \atop {x^2+y^2=-xy+12}} \right.\; \; \left \{ {{-x-y+18=-xy+12} \atop {x^2+y^2=-xy+12}} \right. \\\\\left \{ {{x+y=xy+6} \atop {x^2+y^2=-xy+11}} \right.\; \; \; \; u=x+y\; ,\; \; v=xy\; ,\\\\u^2=x^2+y^2+2xy=x^2+y^2+2v\; \; \to \; \; x^2+y^2=u^2-2v\\\\\left \{ {{u=v+6\qquad } \atop {u^2-2v=-v+12}} \right.\; \; \left \{ {{v=u-6} \atop {u^2-v=12}} \right.\; \;  \left \{ {{v=u-6\qquad } \atop {u^2-(u-6)=12}} \right.\; \; \left \{ {{u=v-6\; \; } \atop {u^2-u-6=0}} \right.$

$u^2-u-6=0\; \; ,\; \; u_1=-2\; ,\; \; u_2=3\; \; (teorema\; Vieta)\\\\v_1=-2-6=-8\; ,\; \; v_2=3-6=-3\\\\a)\; \; \left \{ {{x+y=-2} \atop {xy=-8}} \right.\; \; \left \{ {{y=-x-2} \atop {-x^2-2x+8=0}} \right.\; \; \left \{ {{y=-x-2} \atop {x_1=-4\; ,\; x_2=2}} \right.\; \; \left \{ {{y_1=2\; ,\; y_2=-4} \atop {x_1=-4\; ,\; x_2=2}} \right.\\\\(-4,2)\; ,\; (2,-4)$

$b)\; \; \left \{ {{x+y=3} \atop {xy=-3}} \right. \; \; \left \{ {{y=3-x\; \; \; \; } \atop {3x-x^2=-3}} \right.\; \;\left \{ {{y=3-x\qquad } \atop {x^2-3x-3=0}} \right.\; \; \left \{ {{y=3-x\qquad } \atop {x_{1,2}=\frac{3\pm \sqrt{21}}{6}}} \right. \\\\y_1=3-\frac{3-\sqrt{21}}{6}=\frac{15-\sqrt{21}}{6}\; \; ,\; \; y_2=3-\frac{3+\sqrt{21}}{6}=\frac{15+\sqrt{21}}{6}\\\\(\frac{3-\sqrt{21}}{6}\, ;\, \frac{15-\sqrt{21}}{6}\, )\; \; ,\; \; (\frac{3+\sqrt{21}}{6}\, ;\, \frac{15+\sqrt{21}}{6}\, )\\\\Otvet:\; \; (-4;2)\; ,\; (2;-4)\; ,\;  (\frac{3-\sqrt{21}}{6}\, ;\, \frac{15-\sqrt{21}}{6}\, )\; ,\; (\frac{3+\sqrt{21}}{6}\, ;\, \frac{15+\sqrt{21}}{6}\, )\; .$