Question

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

Answer 1

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

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