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Zarina1111111
@Zarina1111111
July 2022
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решите систему уравнения:
{x^2+y^2+x+y=32
{x^2+y^2-3x-3y=4
A(3;4),(4;3)
B(1,5;2,5),(3,5;-3,5)
C(-1;-2),(-2;-1)
D(-3;-4),(-4;-3)
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Ellenochka
[tex] \left \{ {{x^2+y^2+x+y=32} \atop {x^2+y^2-3x-3y=4}} \right.[/tex] ⇔ [tex]\left \{ {{x^2+y^2=32-(x+y)} \atop {x^2+y^2-3(x+y)=4}} \right. [/tex][tex]\left \{ {{x^2+y^2=32-(x+y)} \atop {32-(x+y)-3(x+y)=4}} \right. [/tex]32-(x+y)-3(x+y)=432-4(x+y)=44(x+y)=28x+y=7[tex] \left \{ {{x^2+y^2=25} \atop {x+y=7}} \right. [/tex] ⇔ [tex] \left \{ {{x^2+(7-x)^2=25} \atop {y=7-x}} \right. [/tex] x²+(7-x)²=25x²+49-14x+x²=252x²-14x+24=0D=196-4*2*24=4=2²x₁=3 u x₂=4[tex] \left \{ {{y_1=4} \atop {x_1=3}} \right. \left \{ {{y_2=3} \atop {x_2=4}} \right. [/tex]Ответ A(3;4),(4;3)
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