Question

Решите систему уравнений:

Answer 1

Заметим, что 17 = 16 + 1 ⇒ 2y² - 4xy + 3x² = y² - x² + 1 ⇔ y² - 4xy + 4x² = 1 ⇔ (y - 2x)² = 1 ⇒ y = 2x + 1 или y = 2x - 1.

$\left \{ {{y=2x+1} \atop {(2x+1)^2-x^2=16}} \right. \\ \left \{ {{y=2x+1} \atop {3x^2+4x-15=0}} \right. \\\left \{ {{y=2x+1} \atop {x=-3; \frac{5}{3}}} \right. \\ \left \{ {{x=-3} \atop {y=-5}} \right. \left \{ {{x=\frac{5}{3}} \atop {y=\frac{13}{3}}} \right.$ или $\left \{ {{y=2x-1} \atop {(2x-1)^2-x^2=16}} \right. \\\left \{ {{y=2x-1} \atop {3x^2-4x-15=0}} \right. \\ \left \{ {{y=2x-1} \atop {x=-\frac{5}{3}; 3}} \right. \\ \left \{ {{x=-\frac{5}{3}} \atop {y=-\frac{13}{3}}} \right. \left \{ {{x=3} \atop {y=5}} \right.$

Ответ: $(-3; -5), (-\frac{5}{3}; -\frac{13}{3}), (\frac{5}{3}; \frac{13}{3}), (3; 5)$