Question

Помогите пожалуйста решить все уравнения и задачу даю 100 баллов. Решения желательно на листке ​

Answer 1

Объяснение:

1)

$\frac{2x {}^{2} + 3x + 1 }{(x + 1)(x - 3) } = 1 \\ \frac{2 {x}^{2} + 2x + x + 1 }{(x + 1)(x - 3)} = \\ \frac{2x(x + 1) + x + 1}{(x + 1)(x - 3)} = \\ \frac{(x + 1)(2x + 1)}{(x + 1)(x - 3)} = \frac{2x + 1}{x - 3}$

2)

$\frac{3}{x - 1} - \frac{3}{x + 3} = 1 \\ \frac{3}{x - 1} - \frac{3}{x + 3} - 1 = 0 \\ \frac{3(x + 3) - 3(x - 1) - (x - 1)(x + 3)}{(x - 1)(x + 3)} = 0 \\ \frac{3x + 9 - 3x + 3(x {}^{2} + 3x - x - 3)}{(x - 1)(x + 3)} \\ = 0 \\ \frac{9 + 3 - {x}^{2} - 2x + 3 }{(x - 1)(x + 3)} = 0 \\ \frac{15 - x {}^{2} - 2x}{(x - 1)(x + 3)} = 0 \\ 15 - x {}^{2} - 2x = 0 \\ - {x}^{2} - 2x + 15 = 0 \\ {x}^{2} + 2x - 15 = 0 \\ d = 4 + 60 = 64 \\ \\ x(1.2) = \frac{ - 2 + - \sqrt{64} }{2 \times 1} = \frac{ - 2 + - 8}{2} \\ x(1) = \frac{ - 2 + 8}{2} = 3 \\ x(2) = \frac{ - 2 - 8}{2} = - 5$

3)

$\frac{x - 1}{x + 2} + \frac{x + 1}{x - 2} + \frac{2x + 8}{4 - {x}^{2} } = 0 \\ \frac{x - 1}{x + 2} + \frac{x + 1}{x - 2} + \frac{2x + 8}{(2 - x)(2 + x)} = 0 \\ \frac{x - 1}{x + 2} + \frac{x + 1}{x - 2} + \frac{2x + 8}{ - (x - 2)(2 + x)} \\ = 0 \\ \frac{x - 1}{x + 2} + \frac{x + 1}{x - 2} - \frac{2x + 8}{(x - 2)(2 + x)} = 0 \\ \frac{(x - 2)(x - 1) + (x + 2)(x + 1) (2x + 8)}{(x -2)(x + 2)} = 0 \\ \frac{ {x}^{2} - x - 2x + 2 + {x}^{2} + x + 2x + 2 - 2x - 8 }{(x - 2)(x + 2)} = 0 \\ \frac{2 {x}^{2} - 4 - 2x }{(x - 2)(x + 2)} = 0 \\ \frac{2( {x}^{2} - 2 - x }{(x - 2)(x + 2)} = 0 \\ \frac{2( {x}^{2} - x - 2)}{(x - 2)(x + 2)} = 0 \\ \frac{2( {x}^{2} + x - 2x - 2) }{(x - 2)(x + 2)} = 0 \\ \frac{2(x(x + 1) - 2(x + 1))}{(x - 2)(x + 2)} = 0 \\ \frac{2(x + 1)(x - 2)}{(x - 2)(x + 2)} = 0 \\ \frac{2(x + 1)}{x + 2} = 0 \\ \frac{2(x + 1)}{x + 2} = 0 \\ 2(x + 1) = 0 \\ x + 1 = 0 \\ x = - 1$