Помогите решить уравнение 1/(x-2)^2-1/x(x-4)=4/3

1/(x^2-4x+4)-1/(x^2-4x) = 4/3
x^2-4x=t
1/(t+4)-1/t=4/3
3t-3(t+4)=4t(t+4)
4t^2+16t+12=0
t^2+4t+3=0
t=-3 --> x^2-4x=-3
x=1
x=3
t=-1-->x^2-4x+1=0
x=2±√3
(≠0, 2, 4)

1/(x-2)^2 - 1/x(x-4) = 4/3

(x - 2) = a
x = a + 2
(x - 4) = (a + 2) - 4 = (a - 2)
=>
1/a^2 - 1/(a+2)*(a-2) = 4/3

1/a^2 - 1/(a^2 - 4) = 4/3

3*(a^2 - 4) - 3a^2 = 4a^2*(a^2 - 4)

3a^2 - 12 - 3a^2 = 4a^4 - 16a^2

4a^4 - 16a^2 + 12 = 0

a^4 - 4a^2 + 3 = 0 ----> a^2 = t

t^2 - 4t + 3 = 0
t(1,2) = [4 + - 2]/2 = 2 + - 1
t1 = 2 - 1 = 1
t2 = 2 + 1 = 3

(a1)^2 = t1 = 1 --------> a1 = -1; a2 = +1
(a2)^2 = t2 = 3 --------> a3 = - V3; a4 = + V3

x = a + 2
x1 = a1 + 2 = (-1) + 2 = 1
x2 = a2 + 2 = 1+ 2 = 3
x3 = a3 + 2 = - V3 + 2
x4 = a4 + 2 = V3 + 2

Проверка при x1 = 1
1/(x-2)^2 - 1/x(x-4) = 4/3
1/(1-2)^2 - 1/1*(1-4) = 4/3
1 + 1/3 = 4/3
4/3 = 4/3