Question

Решите систему уравнений, пожалуйста, дам 20 баллов!!!

Answer 1

$\left \{ {{(x-y)^2+2(x-y)=3} \atop {3x^2+xy+y^2=27}} \right.\\\\(x-y)^2+2(x-y)-3=0\; ,\; \; t^2+2t-3=0\; \; \to \; \; t_1=-3\; ,\; t_2=1\\\\a)\; \; \left \{ {{x-y=-3\qquad } \atop {3x^2+xy+y^2=27}} \right.\; \; \left \{ {{y=x+3\qquad \qquad \qquad } \atop {3x^2+x(x+3)+(x+3)^2=27}} \right.\\\\3x^2+x^2+3x+x^2+6x+9-27=0\\\\5x^2+8x-18=0\; ,\; \; D/4=16+90=106\; ,\; \; x_{1,2}=\frac{-4\pm \sqrt{106}}{5}$

$y_1=\frac{-4-\sqrt{106}}{5}+3=\frac{11-\sqrt{106}}{5}\; \; ,\; \; y_2=\frac{-4+\sqrt{106}}{5}+3=\frac{11+\sqrt{106}}{5}\\\\b)\; \; \left \{ {{x-y=1\qquad } \atop {3x^2+xy+y^2=27}} \right.\; \; \left \{ {{y=x-1\qquad \qquad \qquad } \atop {3x^2+x(x-1)+(x-1)^2=27}} \right.\\\\3x^2+x^2-x+x^2-2x+1-27=0\\\\5x^2-3x-26=0\; \; ,\; \; D=9+520=529\; ,\\\\x_1=\frac{3-23}{10}=-2\; ,\; \; x_2=\frac{3+23}{10}=2,6\\\\y_1=-2-1=-3\; \; ,\; \; y_2=2,6-1=1,6$

$Otvet:\; \; (\frac{-4-\sqrt{106}}{5};\frac{11-\sqrt{106}}{5})\; ,\; (\frac{-4+\sqrt{106}}{5};\frac{11+\sqrt{106}}{5})\; ,\; (-2;-3)\; ,\; (2,6\, ;\, 1,6)\; .$