Question

Срочно, 8 класс алгебра
даю 35 баллов

Answer 1

1)

$\frac{3x + 4}{ {x}^{2} - 16 } = \frac{ {x}^{2} }{{x}^{2} - 16} \\ 3x + 4 = {x}^{2} \\ 3x + 4 - {x}^{2} = 0 \\ - {x}^{2} + 3x + 4 = 0 \\ {x}^{2} - 3x - 4 = 0 \\ {x}^{2} + x - 4x - 4 = 0 \\ x \times (x + 1) - 4(x + 1) = 0 \\ (x + 1) \times (x - 4) = 0 \\ x + 1 = 0 \\ x - 4 = 0 \\ x = - 1$

2)

$5(x - 3) ^{4} - 4(x - {3})^{2} - 1 = 0 \\ 5 {t}^{2} - 4t - 1 = 0 \\ t = - \frac{1}{5} \\ t = 1 \\ (x - {3})^{2} = - \frac{1}{5} \\ (x - {3})^{2} = 1 \\ x = 2 \\ x = 4 \\ \\ x_{1}2 \: \: \: . \: \: x_{2} = 4$

3)

$\frac{3}{x - 5} + \frac{8}{x} = 2 \\ \frac{3}{x - 5} + \frac{8}{x} - 2 = 0 \\ \frac{3x + 8(x - 5) - 2x \times (x - 5)}{x \times (x - 5)} = 0 \\ \frac{3x + 8x - 40 - 2 {x}^{2} + 10x }{x \times (x - 5)} = 0 \\ \frac{21x - 40 - 2 {x}^{2} }{x \times (x - 5)} = 0 \\ 21x - 40 - 2 {x}^{2} = 0 \\ - 2 {x}^{2} + 21x - 40 = 0 \\ 2 {x}^{2} - 21x + 40 = 0 \\ 2 {x}^{2} + 5x - 16x + 40 = 0 \\ x \times (2x - 5) - 8(2x - 5) = 0 \\ (2x - 5) \times (x - 8) = 0 \\ 2x - 5 = 0 \\ x - 8 = 0 \\ \\ x_{1} = \frac{5}{2} \: \: \: \: \: . \: \: \: \: x_{2} = 8$

надеюсь помогла))))