Решить систему уравнений с двумя переменными[tex] \left \{ {{ x^{2} + y^{2} + xy = 13 } \atop {x + y + xy = 7}} \right. [/tex]. Created by qwerty079. algebra-ru

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$\left \{ {{x^2+y^2+xy=13} \atop {x+y+xy=7}} \right. \; ,\; \; x^2+y^2=(x+y)^2-2xy\; ,\; \left \{ {{(x+y)^2-xy=13} \atop {x+y+xy=7}} \right. \\\\u=x+y\; ,\; \; v=xy\; \; \Rightarrow \; \; \left \{ {{u^2-v=13} \atop {u+v=7}} \right. \oplus \; \left \{ {{u^2+u=20} \atop {v=7-u}} \right. \; \left \{ {{u^2+u-20=0} \atop {v=7-u}} \right. \\\\ \left \{ {{u_1=-5\; ,\; u_2=4} \atop {v_1=12\; ,\; v_2=3}} \right. \\\\a)\; \; \left \{ {{x+y=-5} \atop {xy=12}} \right. \; \left \{ {{y=-x-5} \atop {x(-x-5)=12}} \right. \; \left \{ {{y=-x-5} \atop {x^2+5x+12=0}} \right. \; ,\; D=25-4\cdot 12=-23\ \textless \ 0$
нет действительных решений

$b)\; \; \left \{ {{x+y=4} \atop {xy=3}} \right. \; \left \{ {{y=4-x} \atop {x(4-x)=3}} \right. \; \left \{ {{y=4-x} \atop {x^2-4x+3=0}} \right. \; \left \{ {{y_1=3\; ,\; y_2=1} \atop {x_1=1\; ,\; x_2=3}} \right. \\\\Otvet:\; \; (1,3)\; ,\; \; (3,1)\; .$

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sangers X²+y²+xy=13
x+y+xy=7
Суммируем эти уравнения:
x²+2xy+y²+x+y=20
(x+y)²+(x+y)=20=0
Пусть (x+y)=t  ⇒
t²+t-20=0 D=81
t₁=-5
x+y=-5 y=-5-x
-5+xy=7 xy=12 x*(-5-x)=-5x-x²=12 x²+5x+12=0 D=-23 ∉
t₂=4
x+y=4 y=4-x
4+xy=7 xy=3 x*(4-x)=4x-x²=3 x²-4x+3=0 D=4 x₁=3 x₂=1
y₁=4-3=1
y₂=4-1=3
Ответ: x₁=3 y₁=1 x₂=1 y₂=3.

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