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