x > 0 . Let √x = y>0 (*) => the given equation is equivalent to y²/4 - 12/(y²+4y) =1 y²(y²+4y)-12•4 = 4(y²+4y) y⁴+4y³-4y²-16y-48=0 (1) But y⁴+4y³-4y-16y-48= =(y²+2y-12)(y²+2y+4) and the (1) written (y²+2y-12)(y²+2y+4)=0 => y²+2y-12=0 (2), { y²+2y+4=(y+1)²+3>0}. (2) => y=-1±√13 => y=-1+√13, due to (*). From (*) √x = (-1+√13) => x =(-1+√13)² = 14-2√13 accepted because 14-2√13>0.
Απιθανος τροπος. Μπραβο!!!
x > 0 .
Let √x = y>0 (*) => the given equation is equivalent to y²/4 - 12/(y²+4y) =1
y²(y²+4y)-12•4 = 4(y²+4y)
y⁴+4y³-4y²-16y-48=0 (1)
But y⁴+4y³-4y-16y-48=
=(y²+2y-12)(y²+2y+4) and the (1) written (y²+2y-12)(y²+2y+4)=0 =>
y²+2y-12=0 (2), { y²+2y+4=(y+1)²+3>0}.
(2) => y=-1±√13 => y=-1+√13, due to (*).
From (*) √x = (-1+√13) =>
x =(-1+√13)² = 14-2√13 accepted because 14-2√13>0.
Let x= t^2 solving eqn gives
t= +-√13-(1). Valid +ve only for real x= t^ 2= (14)-2√13 soln.
x/2^2 12/x+2^2+x x/1^1+1^1/x+1^2+x (x ➖ 2x+1).