Помогите пожалуйста решить 6.003;6.004;6.005;6.008

6.003) Делаем эту замену и получаем:
z + 3/z + 4 = 0
z^2 + 4z + 3 = 0
(z + 1)(z + 3) = 0
1) z = (x^2+x-5)/x = -1
x^2 + x - 5 = -x
x^2 + 2x - 5 = 0
D = 4 + 4*5 = 24 = (2√6)^2
x1 = (-2 - 2√6)/2 = -1 - √6; x2 = -1 + √6
2) z = (x^2+x-5)/x = -3
x^2 + x - 5 = -3x
x^2 + 4x - 5 = 0
(x + 5)(x - 1) = 0
x3 = -5; x4 = 1

6.004) $x^4-14= \frac{50}{2x^4-7}$
Умножаем все на 2
$2x^4-28= \frac{100}{2x^4-7}$
Замена 2x^4 - 7 = z
z - 21 = 100/z
z^2 - 21z - 100 = 0
(z -25)(z + 4) = 0
1) z = 2x^4 - 7 = -4
2x^4 = 3; x^4 = 1,5
$x1=- \sqrt[4]{1,5} ; x2= \sqrt[4]{1,5}$
2) z = 2x^4 - 7 = 25
2x^4 = 32
x3 = -2; x4 = 2

6.005) $\frac{1}{x^2+2x}- \frac{1}{x^2+2x+1} = \frac{1}{12}$
Замена x^2 + 2x = z
1/z - 1/(z+1) = 1/12
Умножаем на 12z(z+1)
12(z + 1) - 12z = z(z + 1)
12z + 12 - 12z = z^2 + z
z^2 + z - 12 = 0
(z + 4)(z - 3) = 0
1) z = x^2 + 2x = -4
x^2 + 2x + 4 = 0
Решений нет
2) z = x^2 + 2x = 3
x^2 + 2x - 3 = 0
(x + 3)(x - 1) = 0
x1 = -3; x2 = 1

6.008. $\frac{x-3}{x-1}+ \frac{x+3}{x+1} = \frac{x+6}{x+2} + \frac{x-6}{x-2}$
Умножаем все на (x-1)(x+1)(x-2)(x+2) = (x^2-1)(x^2-4)
(x^2-4)*[ (x-3)(x+1) + (x+3)(x-1) ] = (x^2-1)*[ (x+6)(x-2) + (x-6)(x+2) ]
(x^2-4)*(x^2-3x+x-3+x^2+3x-x-3) = (x^2-1)*(x^2+6x-2x-12+x^2-6x+2x-12)
(x^2-4)(2x^2-6) = (x^2-1)(2x^2-24)
2x^4 - 8x^2 - 6x^2 + 24 = 2x^4 - 2x^2 - 24x^2 + 24
-14x^2 = -26x^2
12x^2 = 0
x = 0