Question

xy=10 xy2+x2y=90 vot sistema uravneniy pomoqite

Answer 1

$left { {{xy=10} atop {x^2+y^2=90}} right. ; left { {{2xy=20} atop {x^2+y^2=90}} right. ; left { {{2xy=20} atop {x^2+2xy+y^2=90+10}} right. ; left { {{xy=10} atop {(x+y)^2}-10^2=0} right. ;$
$left { {{xy=10} atop {[(x+y)-10]*[(x+y)+10]}=0} right. ; left { {{xy=10} atop {x=10-y,or,x=-10-y}} right.$

1) $left { {{xy=10} atop {x=10-y}} right. ; left { {{y(10-y)=10} atop {x=10-y}} ; left { {{y^2-10y+10=0} atop {x=10-y}} right. right. ;D=100-40=60=(2 sqrt{15}) ^2$
$y_{1,2}=5+/- sqrt{15};$
$x_{1,2}=15-/+ sqrt{15}$

Две точки: $(5-/+ sqrt{15};5+/- sqrt{15})$

2) $left { {{xy=10} atop {x=-10-y}} right. ; left { {{y(-10-y)=10} atop {x=-10-y}} ; left { {{y^2+10y+10=0} atop {x=-10-y}} right. right. ;D=100-40=60=(2 sqrt{15}) ^2$
$y_{1,2}=-5+/- sqrt{15};$
$x_{1,2}=15-/+ sqrt{15}$

Еще две точки: $(-5-/+ sqrt{15};15+/- sqrt{15})$

Ответ: $(5- sqrt{15};5+ sqrt{15})$,
$(5+ sqrt{15};5- sqrt{15})$,
$(-5- sqrt{15};15+ sqrt{15})$,
$(-5+ sqrt{15};15- sqrt{15})$.