Помогите решить хоть какой из этих номеров

3(6).

а)
(6х+1)(6х-1)(12х-1)(4х+1) = 56,
(6х+1)(12х-1)*(6х-1)(4х+1) = 56,
(72х+12х-6х-1)*(24х+6х-4х-1) = 56,
(72х+6х-1)*(24х+2х-1) = 56,
(6*(12х+х)-1)*(2*(12х+х)-1) = 56,

пусть (12х+х) = а, тогда:

(6а-1)(2а-1) = 56,
12а-6а-2а+1 = 56,
12а-8а+1-56 = 0,
12а-8а-55 = 0,
Д = (-8) - 4*12*(-55) = 64 + 2640 = 2704,
а1= (8 + 2704) / 2*12 = (8 + 52) / 24 = 60/24 = 5/2 = 2,5,
а2= (8 - 2704) / 2*12 = (8 - 52) / 24 = -44/24 = -11/6 = -1 5/6 lt; 0 - не подх,

при а = 2,5:
12х+х = 2,5,
12х+х-2,5 = 0,
Д = 1 - 4*12*(-2,5) = 1 + 120 = 121,
х1 = (-1 + 121) / 2*12 = (-1+11)/24 = 10/24 = 5/12,
х2 = (-1 - 121) / 2*12 = (-1-11)/24 = -12/24 = -1/2,

б)
(6х+1)(6х-1)(12х-1)(3х+1) = -18х,
(6х+1)(6х-1)*(12х-1)(3х+1) = -18х,
(36х-1)*(36х+12х-3х-1) = -18х,
(36х-1)*(36х+9х-1) = -18х,
(36х-1)/х * (36х+9х-1)/х = -18х/х,
(36х/х - 1/х) * (36х/х + 9х/х - 1/х) = -18х/х,
(36х - 1/х) * (36х + 9 - 1/х) = -18,

пусть (36х-1/х) = а, тогда:

а * (а + 9) = -18,
а + 9а + 18 = 0,
Д = 9 - 4*1*18 = 81 - 72 = 9,
а1 = (-9 + 9) / 2*1 = (-9+3) / 2 = -6/2 = -3,
а2 = (-9 - 9) / 2*1 = (-9-3) / 2 = -12/2 = -6,

при а1 = -3:
36х - 1/х = -3,
36х - 1 = -3х,
36х + 3х - 1 = 0,
Д = 3 - 4*36*(-1) = 9 + 144 = 153,
х1 = (-3 + 153) / 2*36 = (-3 + 317) / 72 = (-1+17)/24,
х2 = (-3 - 153) / 2*36 = (-3 - 317) / 72 = (-1-17)/24,

при а2 = -6:
36х - 1/х = -6,
36х - 1 = -6х,
36х + 6х - 1 = 0,
Д = 6 - 4*36*(-1) = 36 + 144 = 180,
х3 = (-6 + 180) / 2*36 = (-6 + 65) / 72 = (-1+5)/12,
х4 = (-6 - 180) / 2*36 = (-6 - 65) / 72 = (-1-5)/12