Решите систему уравнения:{x^2+y^2+x+y=32{x^2+y^2-3x-3y=4A(3;4),(4;3)B(1,5;2,5),(3,5;-3,5)C(-1;-2),(-2;-1)D(-3;-4),(-4;-3)

[latex] \left \{ {{x^2+y^2+x+y=32} \atop {x^2+y^2-3x-3y=4}} \right.[/latex] ⇔ [latex]\left \{ {{x^2+y^2=32-(x+y)} \atop {x^2+y^2-3(x+y)=4}} \right. [/latex] [latex]\left \{ {{x^2+y^2=32-(x+y)} \atop {32-(x+y)-3(x+y)=4}} \right. [/latex] 32-(x+y)-3(x+y)=4 32-4(x+y)=4 4(x+y)=28 x+y=7 [latex] \left \{ {{x^2+y^2=25} \atop {x+y=7}} \right. [/latex] ⇔ [latex] \left \{ {{x^2+(7-x)^2=25} \atop {y=7-x}} \right. [/latex]  x²+(7-x)²=25 x²+49-14x+x²=25 2x²-14x+24=0 D=196-4*2*24=4=2² x₁=3   u   x₂=4 [latex] \left \{ {{y_1=4} \atop {x_1=3}} \right. \left \{ {{y_2=3} \atop {x_2=4}} \right. [/latex] Ответ A(3;4),(4;3)