Question

Помогите. ..........................

Answer 1

$left { {{x^2y^2+2xy=3} atop {(x+y)^2-(x+y)=0}} right. \ left { {{x^2y^2+2xy=3} atop {(x+y)(x+y-1)=0}} right. \(xy)^2+2xy=3 \xy=a \a^2+2a-3=0 \D=4+12=16=4^2 \a_1= frac{-2+4}{2} =1 \a_2= frac{-2-4}{2} =-3 \1) left { {{xy=1} atop {x+y=0}} right. \x=-y \-y^2=1 \y^2=-1 \yin varnothing \2) left { {{xy=1} atop {x+y-1=0}} right. \x=1-y \(1-y)*y=1 \-y^2+y=1 \-y^2+y-1=0 \y^2-y+1=0 \D=1-4 textless 0Rightarrow y in varnothing$
$3) left { {{xy=-3} atop {x+y=0}} right. \x=-y \-y^2=-3 \y^2=3 \y_1=sqrt{3} \y_2=-sqrt{3} \x_1=-sqrt{3} \x_2=sqrt{3} \4) left { {{xy=-3} atop {x+y-1=0}} right. \x=1-y \(1-y)*y=-3 \-y^2+y+3=0 \y^2-y-3=0 \D=1+12=13 \y_3= frac{1+sqrt{13}}{2} \y_4=frac{1-sqrt{13}}{2} \x_3=1- frac{1+sqrt{13}}{2} = frac{2-1-sqrt{13}}{2} = frac{1-sqrt{13}}{2} \x_4=1- frac{1-sqrt{13}}{2} = frac{1+sqrt{13}}{2}$
Ответ: $(-sqrt{3};sqrt{3}), (sqrt{3};-sqrt{3}), ( frac{1-sqrt{13}}{2} ; frac{1+sqrt{13}}{2}), ( frac{1+sqrt{13}}{2};frac{1-sqrt{13}}{2})$