Question

Помогите решить пожалуйста .

Answer 1

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:
$3$

Answer 2

Естественно, первый с объяснением в |текст|