Question

ху = 2,
х2 - 2y = -3;
​

Answer 1

Объяснение:

$\left \{ {{xy=2} \atop {x^2-2y=-3\ |*x\neq 0}} \right.\ \ \ \ \left \{ {{xy=2} \atop {x^3-2xy=-3x}} \right. \ \ \ \ \left \{ {{xy=2} \atop {x^3-2*2+3x=0}\right.\\ \ \ \ \ \left \{ {{xy=2} \atop {x^3+3x-4=0}} \right. }}\ \ \ \ \left \{ {{xy=2} \atop {x^3-1+3x-3=0}} \right. \ \ \ \ \left \{ {x{y=2} \atop {(x-1)(x^2+x+1)+3*(x-1)=0}} \right. \ \ \ \ \left \{ {{xy=2} \atop {(x-1)(x^2+x+1+3)=0}} \right.$

$\left\{\begin{array}{ccc}xy=2\\x-1=0\\x^2+x+4=0\end{array}\right\ \ \ \ \left\{\begin{array}{ccc}y=2\\x=1\\D=-15\ .\varnothing\end{array}\right$

Ответ: x=1    y=2.