Question

дам 50 б срочно, нужно решить систему 2х^4=3х^2у+20 3у^2=2х^2у-5

Answer 1

$left { {{2x^4=3x^2y+20} atop {3y^2=2x^2y-5}} right. \ \ x^2=frac{3y^2+5}{2y} \ \2*( frac{3y^2+5}{2y} )^2=3*frac{3y^2+5}{2y} *y+20\ \ frac{(3y^2+5)^2}{2y^2} =frac{(9y^2+15)*y^2}{2y^2} +frac{40y^2}{2y2} \ \9y^4+30y^2+25=9y^4+15y^2+40y^2$

$25y^2=25\ \ y^2=1\ \ y=pm1\ \ 1)y=1\ \ left { {{2x^4=3x^2+20} atop {2x^2=8}} right. \  left { {{2x^4=3x^2+20} atop {x^2=4}} right. \ left { {{2x^4=3x^2+20} atop {x=pm2}} right. \\ 2)y=-1\ \  left { {{2x^4=-3x^2+20} atop {2x^2=-8}} right. \net  kornej\ \ OTVET:(-2;1) (2;1)$