Question

помогите пожалуйста ​

Answer 1

Объяснение:

a)

$\left \{ {{xy+2x+2y=5} \atop {x^2+y^2+3x+3y=8}} \right.\ \ \ \ \left \{ {{xy+2*(x+y)=5} \atop {x^2+2xy+y^2+3*(x+y)=8+2xy}} \right.\ \ \ \ \left \{ {{xy+2*(x+y)} \atop {(x+y)^2+3*(x+y)=8}+2xy} \right. .$

Пусть: х+у=t         х*у=v      ⇒

$\left \{ {{v+2t=5} \atop {t^2+3t=8+2v}} \right.\ \ \ \ \left \{ {{v=5-2t} \atop {t^2+3t=8+2*(5-2t)}} \right.\ \ \ \ \left \{ {{v=5-2t} \atop {t^2+3t=8+10-4t}} \right. \ \ \ \ \left \{ {{v=5-2t} \atop {t^2+7t-18=0}} \right.$

$\left \{ {{v=5-2t} \atop {D=121\ \ \ \ \sqrt{D} =11}} \right. \ \ \ \ \left \{ {{v_1=23\ \ \ \ v_2=1} \atop {t_1=-9\ \ \ \ t_2=2}} \right. .$

1)

$\left \{ {{x+y=-9} \atop {xy=23}} \right.\ \ \ \ \left \{ {{y=-x-9} \atop {(-x-9)*x=23}} \right. \ \ \ \ \left \{ {{y=-x-9} \atop {-x^2-9x-23=0\ |*(-1)}} \right. \ \ \ \ \left \{ {{y=-x-9} \atop {x^2}+9x+23=0} \right.\ \ \ \ \left \{ {{x\in\varnothing\ \ y\in\varnothing} \atop {D=-11}} \right. .$

2)

$\left \{ {{x+y=2} \atop {x*y=1}} \right.\ \ \ \ \left \{ {{y=2-x} \atop {x*(2-x)=1}} \right. \ \ \ \ \left \{ {{y=2-x} \atop {2x-x^2-1=0\ |*(-1)}} \right. \ \ \ \ \ \left \{ {{y=2-x} \atop {x^2-2x+1=0}} \right. \ \ \ \ \left \{ {{y=2-x} \atop {(x-1)^2=0}} \right.$

$\left \{ {{y=2-x} \atop {x-1=0}} \right.\ \ \ \ \left \{ {{y=1} \atop {x=1}} \right. .$

Ответ: (1;1).

б)

ОДЗ: y≠0.

$\left \{ {{x+\frac{1}{y} =4} \atop {x^2+\frac{x}{y} +\frac{1}{y^2} =13}} \right. \ \ \ \ \left \{ {{xy+1=4y} \atop {x^2y^2+xy+1=13y^2}} \right.\ \ \ \ \left \{ {{xy=4y-1} \atop {(4y-1)^2}+4y-1+1=13y^2} \right. \ \ \ \ \left \{ {{xy=4y-1} \atop {16y^2-8y+1+4y+1=13y^2}} \right.$

$\left \{ {{xy=4y-1} \atop {3y^2-4y+1=0}} \right.\ \ \ \ \left \{ {{xy=4y-1} \atop {D=4\ \ \ \ \sqrt{D}=2 }} \right. \ \ \ \ \left \{ {{xy=4y-1} \atop {y_1=\frac{1}{3}\ \ \ \ y_2= 1 }} \right.\ \ \ \ \left \{ {{x_1=1\ \ \ \ x_2=3} \atop {y_1=\frac{1}{3}\ \ \ \ y_2=1 }} \right. .$

Ответ: (1;1/3),  (3;1).