Помогите пожалуйста решить

1)
(3/x) - (3/[x + 4]) = 1
3 (x + 4) - 3 x = x (x + 4)
12 = x^2 + 4 x
x^2 + 4 x + 4 = 16
(x + 2)^2 = 16
x + 2 = (+/-) 4
x = - 2 (+/-) 4
x = - 6 и x = 2
2)
x / (2 x + 3) = 1 / x
x^2 = 2 x + 3
x^2 - 2 x + 1 = 4
(x - 1)^2 = 4
x - 1 = (+/-) 2
x = 1 (+/-) 2
x = 3 и x = - 1
3)
x^2 + 2 x - 15 = 0
x^2 + 2 x + 1 = 16
(x + 1)^2 = 16
x + 1 = (+/-) 4
x = - 1 (+/-) 4
x = - 5 и x = 3
4)
5 x^2 - 8 x - 4 = 0
D = 144 = 12^2
x = (8 [+/-] 12) / 10
x = - 0.4 и x = 2
5)
x (x - 5) = - 4
([x - 5/2] + 5/2) ([x - 5/2] - 5/2) = - 4
[x - 5/2]^2 - 25/4 = - 4
[x - 5/2]^2 = 9/4
x - 5/2 = (+/-) 3/2
x = (5 [+/-] 3) / 2
x = 1 и x = 4