Question

система уравнении x+y-xy=1; x²+y²-xy=3

Answer 1

$left { {{x+y=1+xy} atop {(x^2+y^2+2xy)-2xy-xy=3}} right. ; left { {{x+y=1+xy} atop {(x+y)^2=3xy+3}} right. ; left { {{x+y=1+xy} atop {(x+y)^2=3(xy+1)}} right. ; \\left { {{x+y=1+xy} atop {(x+y)^2=3(x+y)}} right. ; to ; (x+y)^2-3(x+y)=0,; (x+y)(x+y-3)=0\\1); left { {{x+y=0} atop {x+y=1+xy}} right. ; left { {{x=-y} atop {-y+y=1-y^2}} right. ; left { {{x=-y} atop {0=1-y^2}} right.; left { {{x=-y} atop {y^2=1}} right.$

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

Answer 2

http://znanija.com/task/9720068

Answer 3

это тоже если можете.спасибо