Question

Решите систему уравнений: (x-2)²-(x+3)²=( y-3)²-(y+2)² и( x+2)²+(x-3)²=2x(x-4)+13y надо 1 0

Answer 1

$\left \{ {{(x-2)^2-(x+3)^2=(y-3)^2-(y+2)^2} \atop {(x+2)^2+(x-3)^2=2x(x-4)+13y}} \right. =\left \{ {{10x-5=-10y+5} \atop {2x^2-2x+13=2x^2-8x+13y}} \right.=\\=\left \{ {{y=\frac{2x+2}{2} } \atop {2x^2-2x+13=2x^2-8x+\frac{13(2x+2)}{2} }} \right. =\left \{ {{y=\frac{2x+2}{2}} \atop {6x+13-\frac{13(2x+2)}{2}}} \right.=\left \{ {{y=\frac{2x+2}{2}} \atop {x=0}} \right. =\\=(x;y)=(0;1)$