Question

Решите систему уравнений!!!
$x^{2} +4x - y^{2} -3y=0 sqrt{x+y/x-y} +3sqrt{x-y/x+y} =4$

Answer 1

x^2+4x-y^2-3y=0
√((x+y)/(x-y))+3*√((x-y)/(x+y)) =4

x^2+4x-y^2-3y=0
√(x^2-y^2)*(2y-4x)=4(y^2-x^2)

x^2+4x-y^2-3y=0
√(x^2-y^2)=(4y^2-4x^2)/(2y-4x)

x^2+4x-y^2-3y=0
(y^2-4xy+4x^2)(x^2-y^2)=(4y^4-8x^2y^2+4x^4)

x^2+4x-y^2-3y=0
-y(y-x)(y+x)(5y-4x)=0

y=0
x=0
x=-4

y=x
x^2+4x-x^2-3x=0
x=0
y=0

y=4x/5
x^2+4x-(16x^2/25)-3*(4x/5)=0
x=-40/9
x=0
y=-32/9
y=0