помогите пожалуйста . Created by iiiiiiaaaaaaaooooo. algebra-ru

Объяснение:a)Ответ: (1;1).б)ОДЗ: y≠0.Ответ: (1;1/3), (3;1).

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помогите пожалуйста

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sangers1959

Объяснение:

$>Пусть: х+у=t х*у=v ⇒<img src=$

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

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

$><img src=$

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).

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