Помогите решить срочно нужно!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
{y^4+x=3 /*3
{2x^2-5x+3y^4=1
{ 3y^4+3x=9
3x+5x-2x^2=8
8x-2x^2=8
-2x^2+8x-8=0
2x^2-8x+8=0
D=64-2*4*8= 0
x=8/4=2
y^4+2=3
y^4=1
y=1
x=2
x+y/x-y +x/y=-5/6
x^2+xy/xy-y^2=1/6
(x+y)/(x-y)+(x/y)=-5/6
(x(x+y)/(y(x-y))=1/6
x/y=1/6
6x=y
x+6x/x-6x+x/6x=-5/6
7x/-5x+6=-5/6
-42x^2+5=-25
-42x^2=-30
7x^2=5
x=V5/7
y=6V5/7
3x^2-xy=10y^2
x^2-2xy+y^2=4 => (x-y)^2=4
x-y=2
x=2+y
3(2+y)^2-(2+y)*y=10y^2
12+12y+3y^2-2y-y^2=10y^2
-8y^2+10y+12=0
8y^2-10y-12=0
D=100-4*8*-12= V484=22
y1=10-22/16=-12/16=-3/4
y2=10+22/16=32/16=2
x=4