Помогите решить пожалуйста .

1)
$\frac{x + 4}{x + 1} - \frac{10}{ {x}^{2} - 1} = \frac{10}{3} ; \\ | {a}^{2} - b {}^{2} = (a - b)(a + b) | \\ \frac{ x + 4}{x + 1} - \frac{10}{(x - 1)(x + 1)} = \frac{10}{3} ; \\ | \frac{a}{b} - \frac{t}{bc} = \frac{ac - t}{bc} | \\ \frac{(x - 4)(x - 1) - 10}{(x - 1)(x + 1)} = \frac{10}{3} ; \\ |(a + b)(c + d) = ac + ad + bc + bd| \\ \frac{ {x { }^{2} - x - 4x + 4 - 10} }{(x - 1)(x + 1)} = \frac{10}{3} ; \\ | \frac{a}{b} = \frac{c}{d} ; \: \: ad =bc | \\ \frac{ {x}^{2} - 5x - 6 }{ {x}^{2} - 1} = \frac{10}{3} ; \\ 3 {x}^{2} - 15x -18 = 10x {}^{2} - 10 ; \\ |a = - b; \: \: a + b = 0| \\ 7 {x}^{2} + 15x + 8 = 0; \\ D = b {}^{2} - 4ac = 225 - 224 = 1 \\ x_{1 } = \frac{ - b + \sqrt{ D} }{2a} = \frac{ - 15 + 1}{14} = - 1 \\ x_{1 } = \frac{ - b - \sqrt{ D} }{2a} = \frac{ - 15 - 1}{14} = - \frac{8}{7} = - 1 \frac{1}{7}$
ОДЗ:
x≠±1
Ответ:
$- 1 \frac{1}{7}$

2)
$\frac{x}{x + 4} + \frac{5}{x - 4} = \frac{32}{ {x}^{2} - 16 } \\ \frac{x(x - 4) + 5(x + 4)}{ {x}^{2} - 16} = \frac{ 32}{ {x}^{2} - 16} \\ \frac{ {x}^{2} - 4x + 5x + 20 - 32 }{ {x}^{2} - 16 } = 0 \\ \frac{ {x}^{2} + x - 12}{ {x}^{2} - 16} = 0 \\ |x_{1} = 3 \\ |x_2 = - 4$
ОДЗ:
х≠±4

Ответ:
$3$