Question

Помогите решить срочно нужно!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

{y^4+x=3 /*3

{2x^2-5x+3y^4=1 

 

{ 3y^4+3x=9

{2x^2-5x+3y^4=1

 

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

3x^2-xy=10y^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