Question

Решите систему уравнений

Answer 1

Ответ:

1) (0,2)(0,-2)(2,0);

2)(1,3)(-1,-3);

3)(0,1)(0,-1)(1,1)(-1,-1).

Пошаговое объяснение:

$1)\\\left \{ {{y^2-x^2=4-4x} \atop {x^2+y^2-3xy=4}} \right. => \left \{ {{y^2=x^2-4x+4} \atop {x^2+x^2-4x+4-3xy=4}} \right. => \left \{ {{y^2=x^2-4x+4} \atop {2x^2-4x-3xy=0}} \right. => \left \{ {{y^2=x^2-4x+4} \atop {x(2x-3y-4) = 0}} \right. => \left \{ {{y^2=x^2-4x+4} \atop {x = 0}} \right. or \left \{ {{y^2=x^2-4x+4} \atop {x=1.5y+2}} \right. \\1.1) \left \{ {{y^2=4} \atop {x = 0}} \right. => \left \{ {{y=2,-2} \atop {x=0}} \right.$

$1.2) \left \{ {{y^2-(1.5y+2)^2+6y+8=4} \atop {x=1.5y+2}} \right. => \left \{ {{y^2-2.25y^2=0} \atop {x=1.5y + 2}} \right. => \left \{ {{y=0} \atop {x=1.5y+2}} \right. => \left \{ {{y=0} \atop {x=2}} \right.$

$2)\\\left \{ {{y^2-2xy-3x^2=0} \atop {y^2-xy-2x^2=4}} \right. => \left \{ {{y^2=3x^2+2xy} \atop {3x^2+2xy-xy-2x^2=4}} \right. => \left \{ {{y^2=3x^2+2xy} \atop {x^2+xy=4}} \right. => \left \{ {{y^2=x^2+8} \atop {x(x+y)=4}} \right. => \left \{ {{(y-x)(y+x)=8} \atop {x+y=\frac{4}{x} }} \right. => \left \{ {{(y-x)*\frac{4}{x} =8} \atop {x+y=\frac{4}{x} }} \right. => \left \{ {{\frac{4y}{x} - 4 =8} \atop {x+y=\frac{4}{x }} \right. => \left \{ {{y=3x} \atop {x=\frac{1}{x} }} \right.$

$=> \left \{ {{y=3,-3} \atop {x=1,-1}} \right.$

$3)\\\left \{ {{2x^2-2xy+3y^2=3} \atop {x^2-xy+2y^2=2}} \right. => \left \{ {{3y^2-4y^2=-1} \atop {x^2-xy+2y^2=2}} \right. => \left \{ {{y=1,-1} \atop {x^2-xy+2y^2=2}} \right. \\3.1) y = 1, x^2-x = 0 => x(x-1) = 0 => x = 0,1\\3.2) y = -1, x^2+x = 0 => x(x+1) = 0 => x = 0,-1$